Chapter 03 Review Problems

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Solutions Compact

Nomenclature and Chemical Formulas

Fill out the missing information regarding a binary ionic compound.

1. Name: sodium chloride
2. Cation:
3. Anion:
4. Formula:

Solution

Answer:

1. Name: sodium chloride
2. Cation: Na⁺
3. Anion: Cl⁻
4. Formula: NaCl

Concept: ionic compound nomenclature and atomic structure

1. Identify the Ions:
   • “Sodium” refers to the sodium ion. As a Group 1 metal, it forms a Na⁺ cation.
   • “Chloride” refers to the chloride ion. As a Group 17 nonmetal, it forms a Cl⁻ anion.
2. Determine the Formula: Ionic compounds must be electrically neutral. The ratio of cations to anions must be such that the total positive charge equals the total negative charge. In this case, one 1+ cation balances one 1− anion. Therefore, the formula is NaCl.

Fill out the missing information regarding a binary ionic compound.

1. Name:
2. Cation: Mg²⁺
3. Anion: O²⁻
4. Formula:

Solution

Answer:

1. Name: magnesium oxide
2. Cation: Mg²⁺
3. Anion: O^2–
4. Formula: MgO

Concept: ionic compound nomenclature and atomic structure

1. Determine the Formula: Ionic compounds must be electrically neutral. We need to find the simplest whole-number ratio of cations to anions that will result in a net charge of zero. In this case, one 2+ cation (Mg²⁺) is perfectly balanced by one 2− anion (O²⁻). Therefore, the formula is MgO.

2. Determine the Name: For a binary ionic compound containing a main-group metal (like Mg), the naming rule is: (Name of Cation) (Base Name of Anion + –ide suffix)

   • The cation Mg²⁺ is simply named magnesium.
   • The anion O²⁻ is named by taking the base name of oxygen (“ox–”) and adding the “–ide” suffix, giving oxide.
   • Combining them gives the name magnesium oxide.

Fill out the missing information regarding a binary ionic compound.

1. Name:
2. Cation:
3. Anion:
4. Formula: Al₂S₃

Solution

Answer:

1. Name: aluminum sulfide
2. Cation: Al³⁺
3. Anion: S^2–
4. Formula: Al₂S₃

Concept: ionic compound nomenclature and atomic structure

1. Deduce the Ions: We know that the total positive charge from the cations must balance the total negative charge from the anions. We can use the subscripts in the formula and the known charges of the elements to find the correct ions.

   Aluminum (Al) is in Group 13 and typically forms a 3+ cation (Al³⁺). Sulfur (S) is in Group 16 and typically forms a 2− anion (S^2–).

   • Charge-Balance Check: (2 × +3) + (3 × −2) = +6 + (−6) = 0. The charges balance.
   • The cation is Al³⁺ and the anion is S^2–.

2. Determine the Name: For a binary ionic compound containing a main-group metal (like Al), the naming rule is: (Name of Cation) (Base Name of Anion + -ide suffix)

   • The cation Al³⁺ is simply named aluminum.
   • The anion S^2– is named by taking the base name of sulfur (“sulf–”) and adding the “–ide” suffix, giving sulfide.
   • Combining them gives the name aluminum sulfide.

Fill out the missing information regarding a binary ionic compound.

1. Name: iron(III) oxide
2. Cation:
3. Anion:
4. Formula:

Solution

Answer:

1. Name: iron(III) oxide
2. Cation: Fe³⁺
3. Anion: O^2–
4. Formula: Fe₂O₃

Concept: ionic compound nomenclature and atomic structure

1. Identify the Ions:
   • Cation: The name “iron(III)” tells us two things. “Iron” is the element Fe. The Roman numeral (III) explicitly states that the iron cation has a charge of 3+. Therefore, the cation is Fe³⁺.
   • Anion: “Oxide” refers to the ion of oxygen. As a Group 16 nonmetal, oxygen forms a 2− anion, O^2–.
2. Determine the Formula: Ionic compounds must be electrically neutral. We need to find the simplest whole-number ratio of cations to anions that will result in a net charge of zero. To balance the total charge, we need to find the least common multiple of the charge magnitudes (3 and 2), which is 6.
   • To get a total positive charge of 6+, we need two Fe³⁺ ions (2 × +3 = +6).
   • To get a total negative charge of 6−, we need three O^2– ions (3 × −2 = −6).
   This gives us the ratio of 2 iron ions to 3 oxide ions. Therefore, the formula is Fe₂O₃.

Fill out the missing information regarding a binary ionic compound.

1. Name:
2. Cation: Sn²⁺
3. Anion: Cl⁻
4. Formula:

Solution

Answer:

1. Name: tin(II) chloride
2. Cation: Sn²⁺
3. Anion: Cl⁻
4. Formula: SnCl₂

Concept: ionic compound nomenclature and atomic structure

1. Determine the Formula: Ionic compounds must be electrically neutral. To balance the 2+ charge of the tin ion, we need two chloride ions, each with a 1− charge (2 × −1 = −2). Therefore, the formula is SnCl₂.

2. Determine the Name: The naming rule for this type of compound is: (Name of Cation)(Roman Numeral for Cation Charge) (Base Name of Anion + -ide suffix)

   • Identify the Cation Type: Tin (Sn) is a metal that can form more than one stable cation (most commonly Sn²⁺ and Sn⁴⁺). Because of this, we must use the Stock system (Roman numerals) to specify its charge.
   • Name the Cation: The cation is Sn²⁺. Its name is tin(II).
   • Name the Anion: The anion Cl^– is named by taking the base name of chlorine (“chlor–”) and adding the “–ide” suffix, giving chloride.
   • Combining them gives the name tin(II) chloride.

Fill out the missing information regarding a binary ionic compound.

1. Name:
2. Cation:
3. Anion:
4. Formula: PbF₄

Solution

Answer:

1. Name: lead(IV) fluoride
2. Cation: Pb⁴⁺
3. Anion: F⁻
4. Formula: PbF₄

Concept: ionic compound nomenclature and atomic structure

1. Deduce the Ions: The key to this problem is to start with the anion, whose charge is predictable from the periodic table.

   • Anion: Fluorine (F) is a halogen in Group 17. It always forms an ion with a 1− charge. The anion is F⁻.
   • Cation: The formula shows there are four F⁻ ions, giving a total negative charge of 4 × (−1) = −4. To ensure the compound is electrically neutral, the single lead (Pb) ion must have a charge of 4+. Therefore, the cation is Pb⁴⁺.

2. Determine the Name: The naming rule for this type of compound is: (Name of Cation)(Roman Numeral for Cation Charge) (Base Name of Anion + -ide suffix)

   • Identify the Cation Type: Lead (Pb) is a metal that can form more than one stable cation. Because of this, we must use the Stock system (Roman numerals) to specify its charge.
   • Name the Cation: The cation is Pb⁴⁺. Its name is lead(IV).
   • Name the Anion: The anion F⁻ is named by taking the base name of fluorine (“fluor–”) and adding the “–ide” suffix, giving fluoride.
   • Combining them gives the name lead(IV) fluoride.

Fill out the missing information regarding an ionic compound.

1. Name: sodium hydroxide
2. Cation:
3. Anion:
4. Formula:

Solution

Answer:

1. Name: sodium hydroxide
2. Cation: Na⁺
3. Anion: OH⁻
4. Formula: NaOH

Concept: ionic compound nomenclature and atomic structure

This problem involves a polyatomic ion. A polyatomic ion is a charged species composed of two or more atoms covalently bonded together that acts as a single ionic unit.

1. Identify the Ions:
   • Cation: “Sodium” refers to the sodium ion. As a Group 1 metal, it forms a Na⁺ cation.
   • Anion: “Hydroxide” is the name of a common polyatomic ion with the formula OH⁻. The names and formulas of common polyatomic ions must be memorized.
2. Determine the Formula: Ionic compounds must be electrically neutral. In this case, one 1+ cation (Na⁺) is perfectly balanced by one 1− polyatomic anion (OH⁻). Therefore, the formula is NaOH.

Fill out the missing information regarding an ionic compound.

1. Name:
2. Cation: Mg²⁺
3. Anion: CO₃²⁻
4. Formula:

Solution

Answer:

1. Name: magnesium carbonate
2. Cation: Mg²⁺
3. Anion: CO₃²⁻
4. Formula: MgCO₃

Concept: ionic compound nomenclature and atomic structure

1. Determine the Formula: Ionic compounds must be electrically neutral. We need to find the simplest whole-number ratio of cations to anions that will result in a net charge of zero. In this case, one 2+ cation (Mg²⁺) is perfectly balanced by one 2− polyatomic anion (CO₃^2–). Therefore, the formula is MgCO₃.

2. Determine the Name: The naming rule for an ionic compound is: (Name of Cation) (Name of Anion)

   • The cation Mg²⁺ is simply named magnesium.
   • The polyatomic anion CO₃²⁻ is named carbonate.
   • Combining them gives the name magnesium carbonate.

Fill out the missing information regarding an ionic compound.

1. Name:
2. Cation:
3. Anion:
4. Formula: Na₂SO₃

Solution

Answer:

1. Name: sodium sulfite
2. Cation: Na⁺
3. Anion: SO₃²⁻
4. Formula: Na₂SO₃

Concept: ionic compound nomenclature and atomic structure

1. Deduce the Ions: The key to this problem is to start with the cation, whose charge is predictable from the periodic table.
   • Cation: Sodium (Na) is a Group 1 metal. It always forms an ion with a 1+ charge. The cation is Na⁺.
   • Anion: The formula shows there are two Na⁺ ions, giving a total positive charge of 2 × (+1) = +2. To ensure the compound is electrically neutral, the single polyatomic anion (SO₃) must have a charge of 2−. Therefore, the anion is the sulfite ion, SO₃²⁻.
2. Determine the Name: The naming rule for an ionic compound is: (Name of Cation) (Name of Anion)
   • The cation Na⁺ is simply named sodium.
   • The polyatomic anion SO₃²⁻ is named sulfite.
   • Combining them gives the name sodium sulfite.

Fill out the missing information regarding an ionic compound.

1. Name:
2. Cation: Fe²⁺
3. Anion: OH⁻
4. Formula:

Solution

Answer:

1. Name: iron(II) hydroxide
2. Cation: Fe²⁺
3. Anion: OH⁻
4. Formula: Fe(OH)₂

Concept: ionic compound nomenclature and atomic structure

1. Determine the Formula: Ionic compounds must be electrically neutral. To balance the 2+ charge of the iron(II) ion, we need two hydroxide ions, each with a 1− charge (2 × −1 = −2).

   Because we need more than one polyatomic ion, we must enclose the ion in parentheses followed by the subscript. This shows that the subscript ‘2’ applies to the entire hydroxide unit (both the O and the H). Therefore, the formula is Fe(OH)₂. (Writing FeOH₂ would be incorrect, as it would imply only the hydrogen atom is doubled.)

2. Determine the Name: The naming rule for this type of compound is: (Name of Cation)(Roman Numeral for Cation Charge) (Name of Anion)

   • Identify the Cation Type: Iron (Fe) is a transition metal that can form more than one stable cation. Therefore, we must use the Stock system (Roman numerals) to specify its charge.
   • Name the Cation: The cation is Fe²⁺. Its name is iron(II).
   • Name the Anion: The polyatomic anion OH⁻ is named hydroxide.
   • Combining them gives the name iron(II) hydroxide.

Fill out the missing information regarding an ionic compound.

1. Name:
2. Cation:
3. Anion:
4. Formula: TiPO₃

Solution

Answer:

1. Name: titanium(III) phosphite
2. Cation: Ti³⁺
3. Anion: PO₃³⁻
4. Formula: TiPO₃

Concept: ionic compound nomenclature and atomic structure

1. Deduce the Ions: The key to this problem is to first identify the polyatomic anion and its charge, which is a matter of memorization.

   • Anion: The polyatomic ion PO₃ is the phosphite ion. It has a charge of 3−. The anion is PO₃³⁻.
   • Cation: The formula shows a 1:1 ratio of the cation to the anion. To ensure the compound is electrically neutral, the single titanium (Ti) ion must have a charge of 3+ to balance the 3− charge of the phosphite ion. Therefore, the cation is Ti³⁺.

2. Determine the Name: The naming rule for this type of compound is: (Name of Cation)(Roman Numeral for Cation Charge) (Name of Anion)

   • Identify the Cation Type: Titanium (Ti) is a transition metal that can form more than one stable cation. Therefore, we must use the Stock system (Roman numerals) to specify its charge.
   • Name the Cation: The cation is Ti³⁺. Its name is titanium(III).
   • Name the Anion: The polyatomic anion PO₃³⁻ is named phosphite.
   • Combining them gives the name titanium(III) phosphite.

What is the proper chemical formula for lithium hypoiodite?

1. LiIO
2. LiO₂
3. LiIO₃
4. LiIO₄

Solution

Answer: A

Concept: nomenclature of ionic compounds and oxyanions

This problem requires understanding the systematic naming of oxyanions (polyatomic anions containing oxygen). For halogens like iodine, there is a series of four related oxyanions. The naming system is based on a reference ion, the “–ate” ion.

1. Identify the Ions:

   • Cation: “Lithium” refers to the lithium ion. As a Group 1 metal, it always forms a Li⁺ cation.
   • Anion: “hypoiodite” is part of the iodine oxyanion series.

2. Determine the Anion Formula from the Name: The system is based on the number of oxygen atoms relative to the “–ate” ion.

   • Periodate (IO₄^–): One more oxygen than the “–ate” ion. (per- prefix)
   • Iodate (IO₃^–): The reference “–ate” ion.
   • Iodite (IO₂^–): One less oxygen than the “–ate” ion. (-ite suffix)
   • Hypoiodite (IO^–): Two less oxygens than the “–ate” ion. (hypo- prefix and -ite suffix)

   The name “hypoiodite” tells us the anion is IO^–.

3. Determine the Compound Formula: To balance the 1+ charge of the Li⁺ cation, we need one 1− hypoiodite anion. Therefore, the formula is LiIO.

Provide the names for the following compounds.

1. N₂O
2. P₂O₅
3. SF₂
4. S₂O
5. P₂I₄

Solution

Answer:

1. dinitrogen monoxide
2. diphosphorus pentoxide
3. sulfur difluoride
4. disulfur monoxide
5. diphosphorus tetraiodide

Concept: nomenclature of covalent compounds

These compounds are all formed between two nonmetals, so they are molecular (covalent) compounds, not ionic. They are named using a system of Greek prefixes to indicate the number of atoms of each element.

The naming rule is: (Prefix for first element)(Name of first element) (Prefix for second element)(Base name of second element + -ide suffix)

• The prefix is usually omitted for the first element if there is only one atom (e.g., SF₂ is “sulfur difluoride,” not “monosulfur difluoride”).
• The o or a at the end of a prefix is often dropped when the element name begins with a vowel (e.g., P₂O₅ is “pentoxide,” not “pentaoxide”).

Which of the following contains both ionic and covalent bonds?

1. CaBr₂
2. COS
3. BaSO₄
4. SF₆
5. none of these

Solution

Answer: C

Concept: ionic and covalent bonding in compounds

The key to this question is to identify an ionic compound that is formed from a polyatomic ion. Polyatomic ions are groups of atoms held together by covalent bonds that carry an overall net charge.

• An ionic bond typically forms between a metal and a nonmetal (or a polyatomic ion).
• A covalent bond typically forms between two nonmetals.

Let’s analyze the options:

• (c) BaSO₄ (barium sulfate): This is the correct answer.
  • It is an ionic compound formed between the metal cation Ba²⁺ and the polyatomic anion SO₄^2– (sulfate). The electrostatic attraction between these two ions is an ionic bond.
  • Within the sulfate anion itself, the central sulfur atom is bonded to four oxygen atoms. Because S and O are both nonmetals, these bonds are covalent.
  • Therefore, BaSO₄ contains both ionic and covalent bonds.

The other options are incorrect:

• (a) CaBr₂ (calcium bromide): This is a simple binary ionic compound formed between a metal (Ca) and a nonmetal (Br). It contains only ionic bonds.
• (b) COS (carbonyl sulfide) and (d) SF₆ (sulfur hexafluoride): These are both molecular compounds formed exclusively from nonmetals. They contain only covalent bonds.

What is the name of C₃H₈?

1. hexane
2. propane
3. decane
4. butane
5. ethane

Solution

Answer: B

Concept: naming simple alkanes

This question requires knowledge of the systematic naming conventions for simple alkanes, which are hydrocarbons (compounds of carbon and hydrogen) containing only single bonds. The name of an alkane is determined by a prefix that indicates the number of carbon atoms, followed by the suffix –ane.

The chemical formula given is C₃H₈. The subscript ‘3’ on the carbon indicates that there are three carbon atoms. According to the naming system, the prefix for three carbons is prop–.

Therefore, the name of the compound is propane.

What is the name of this compound?

Solution

Answer: octanol

Concept: naming simple alcohols

This problem requires a two-step process: first, identify the longest carbon chain (the “parent alkane”), and second, identify the primary functional group and modify the name accordingly.

1. Identify the Parent Alkane: The skeletal structure shows a chain of eight carbon atoms connected by single bonds. An eight-carbon alkane is named octane.

2. Identify the Functional Group and Name the Compound: The molecule contains a hydroxyl (–OH) group, which identifies it as an alcohol. To name an alcohol, we take the name of the parent alkane, drop the final “–e”, and add the suffix –ol.

   octane → octan + -ol → octanol

What is the name of PH₃?

Solution

Answer: phosphine

Concept: nomenclature of covalent compounds

This question addresses an important aspect of chemical nomenclature: some simple molecular compounds have traditional common names that are so widely used they are universally accepted by IUPAC.

1. Systematic Name: Following the standard rules for binary molecular compounds (using prefixes), PH₃ would be named phosphorus trihydride.

2. Common Name: However, this compound is almost exclusively known by its common (or trivial) name, phosphine. This is similar to other simple hydrides like H₂O (water) and NH₃ (ammonia).

While “phosphorus trihydride” is the technically correct systematic name, phosphine is the name used by chemists and is the expected answer.

Write the empirical formulas for the following compounds.

1. iron(II) sulfide
2. iron(III) sulfide
3. iron(II) sulfate
4. iron(III) sulfate
5. iron(II) sulfite

Solution

Answer:

1. FeS
2. Fe₂S₃
3. FeSO₄
4. Fe₂(SO₄)₃
5. FeSO₃

Concept: nomenclature and writing formulas for ionic compounds with multivalent and polyatomic ions

Each problem requires identifying the cation and anion from the name and then combining them in the correct ratio to achieve a neutral compound.

1. iron(II) sulfide:
   • Cation: iron(II) → Fe²⁺
   • Anion: sulfide → S^2–
   • The 2+ and 2− charges balance in a 1:1 ratio. Formula: FeS.
2. iron(III) sulfide:
   • Cation: iron(III) → Fe³⁺
   • Anion: sulfide → S^2–
   • To balance the charges, we need two Fe³⁺ ions (6+ total) and three S^2– ions (6− total). Formula: Fe₂S₃.
3. iron(II) sulfate:
   • Cation: iron(II) → Fe²⁺
   • Anion: sulfate → SO₄^2–
   • The 2+ and 2− charges balance in a 1:1 ratio. Formula: FeSO₄.
4. iron(III) sulfate:
   • Cation: iron(III) → Fe³⁺
   • Anion: sulfate → SO₄^2–
   • To balance the charges, we need two Fe³⁺ ions (6+ total) and three SO₄^2– ions (6− total). Because we need more than one polyatomic ion, we must use parentheses. Formula: Fe₂(SO₄)₃.
5. iron(II) sulfite:
   • Cation: iron(II) → Fe²⁺
   • Anion: sulfite → SO₃^2–
   • The 2+ and 2− charges balance in a 1:1 ratio. Formula: FeSO₃.

Provide the name for each of the following and identify it as a molecular or ionic compound.

Solution

Answer:

Concept: classifying and naming ionic and molecular compounds

1. CH₃OH: Contains only nonmetals (C, H, O), so it is molecular. It is an alcohol with one carbon (methan–) and an –ol suffix: methanol.
2. Ca₃P₂: Contains a metal (Ca) and a nonmetal (P), so it is ionic. Ca (Group 2) is Ca²⁺. P (Group 15) is P³⁻ (phosphide). The name is calcium phosphide.
3. KC₂H₃O₂: Contains a metal (K), so it is ionic. K⁺ is potassium. C₂H₃O₂⁻ is the polyatomic ion acetate. The name is potassium acetate.
4. NaHCO₃: Contains a metal (Na), so it is ionic. Na⁺ is sodium. HCO₃⁻ is the polyatomic ion hydrogen carbonate (or bicarbonate). The name is sodium hydrogen carbonate.
5. FeCl₃: Contains a metal (Fe), so it is ionic. Cl is the chloride ion (Cl⁻). Three Cl⁻ ions give a −3 charge, so the iron must be Fe³⁺. Since iron is a transition metal, we use a Roman numeral. The name is iron(III) chloride.

Provide the systematic name for each of the following and identify its compound type (ionic, molecular, acid, base, or ionic hydrate).

Solution

Answer:

Concept: nomenclature and classification

• HNO₂(aq): This is an acid. The name is derived from the polyatomic anion NO₂⁻ (“nitrite”). The “–ite” suffix for the anion becomes an “–ous acid” suffix. The name is nitrous acid.
• Hg₂O: This is ionic. It contains the uncommon polyatomic cation mercury(I), which has the formula Hg₂²⁺. To balance the O²⁻ anion, one Hg₂²⁺ ion is needed. The name is mercury(I) oxide.
• CuSe: This is ionic. Selenium (Se, Group 16) forms a selenide ion, Se²⁻. To balance this, the copper ion must be Cu²⁺. Since copper is a transition metal, a Roman numeral is required. The name is copper(II) selenide.
• Cl₂O₇: Contains only nonmetals, so it is molecular. We use prefixes: two chlorine (“di–”) and seven oxygen (“hepta–”). The name is dichlorine heptaoxide.
• H₂O₂: This is a molecular compound universally known by its common name, hydrogen peroxide. Its systematic name is dihydrogen dioxide.
• Be(NO₃)₂ · 4 H₂O: This is an ionic hydrate. We name the ionic compound first (Beryllium is Be²⁺, Nitrate is NO₃⁻ → beryllium nitrate). Then we add a prefix for the number of water molecules (“tetra–”) followed by “–hydrate”. The name is beryllium nitrate tetrahydrate.

Provide the systematic name for each of the following and identify its compound type (ionic, molecular, acid, base, or ionic hydrate). Include phase labels where necessary.

Solution

Answer:

Concept: nomenclature and classification

• sulfurous acid: This is an acid. The “–ous acid” suffix corresponds to the “–ite” anion, which is sulfite (SO₃^2–). To balance the charge, two H⁺ ions are needed. Formula: H₂SO₃(aq).
• iron(II) bromide: This is ionic. Iron(II) is Fe²⁺. Bromide is Br⁻. Two bromide ions are needed to balance the charge. Formula: FeBr₂.
• manganese(IV) thiocyanate: This is ionic. Manganese(IV) is Mn⁴⁺. Thiocyanate is the polyatomic anion SCN^–. Four thiocyanate ions are needed, requiring parentheses. Formula: Mn(SCN)₄.
• cadmium phosphite: This is ionic. Cadmium (a transition metal) reliably forms a Cd²⁺ ion. Phosphite is the polyatomic anion PO₃³⁻. To balance the charges (least common multiple is 6), three Cd²⁺ ions and two PO₃³⁻ ions are needed. Formula: Cd₃(PO₃)₂.
• dihydrogen monosulfide: Contains only nonmetals, so it is molecular. The name directly translates to the formula H₂S. (This compound is more commonly known as hydrogen sulfide).
• potassium permanganate: This is ionic. Potassium (K, Group 1) is K⁺. Permanganate is the polyatomic anion MnO₄^–. The charges balance in a 1:1 ratio. Formula: KMnO₄.

Provide the systematic name for each of the following and identify its compound type (ionic, molecular, acid, base, or ionic hydrate). Include phase labels where necessary.

Solution

Answer:

Concept: nomenclature and classification

• iron(II) oxalate: This is ionic. Iron(II) is Fe²⁺. Oxalate is the polyatomic anion C₂O₄²⁻. The charges balance 1:1. Formula: FeC₂O₄.
• sodium chromate: This is ionic. Sodium (Group 1) is Na⁺. Chromate is the polyatomic anion CrO₄²⁻. Two Na⁺ ions are needed. Formula: Na₂CrO₄.
• beryllium dichromate: This is ionic. Beryllium (Group 2) is Be²⁺. Dichromate is the polyatomic anion Cr₂O₇²⁻. The charges balance 1:1. Formula: BeCr₂O₇.
• perchloric acid: This is an acid. The per- prefix and -ic acid suffix correspond to the perchlorate anion, ClO₄⁻. One H⁺ is needed to balance the charge. Formula: HClO₄(aq).
• nitric acid: This is an acid. The -ic acid suffix corresponds to the nitrate anion, NO₃⁻. One H⁺ is needed. Formula: HNO₃(aq).
• sodium carbonate decahydrate: This is an ionic hydrate. The ionic part is sodium (Na⁺) and carbonate (CO₃²⁻) → Na₂CO₃. The prefix “deca–” means ten, and “–hydrate” refers to water molecules. Formula: Na₂CO₃ · 10 H₂O.

Balancing Equations

Write a balanced equation for the following reaction by placing appropriate stoichiometric coefficients.

\[\begin{align*} {\color{currentcolor}\rule[-1.0pt]{2em}{0.5pt}}~\mathrm{CH_3OH} ~+~ {\color{currentcolor}\rule[-1.0pt]{2em}{0.5pt}}~\mathrm{O_2} ~\longrightarrow~ {\color{currentcolor}\rule[-1.0pt]{2em}{0.5pt}}~\mathrm{CO_2} ~+~ {\color{currentcolor}\rule[-1.0pt]{2em}{0.5pt}}~\mathrm{H_2O} \end{align*}\]

Solution

Answer:

\[\begin{align*} \mathrm{2~CH_3OH} + \mathrm{3~O_2} \longrightarrow \mathrm{2~CO_2} + \mathrm{4~H_2O} \end{align*}\]

Concept: balancing equations

A reliable method for balancing the combustion of a compound containing C, H, and O is to balance the elements in the following order: Carbon, then Hydrogen, then Oxygen.

Step 1: Balance Carbon (C) Start with the unbalanced equation. There is 1 C atom on the left and 1 C on the right. The carbons are already balanced. \[ \mathrm{1~CH_3OH} + \mathrm{O_2} \longrightarrow \mathrm{1~CO_2} + \mathrm{H_2O} \]

Step 2: Balance Hydrogen (H) There are 4 H atoms on the left (in 1 CH₃OH). To get 4 H atoms on the right, we need 2 H₂O molecules. \[ \mathrm{CH_3OH} + \mathrm{O_2} \longrightarrow \mathrm{CO_2} + \mathrm{2~H_2O} \]

Step 3: Balance Oxygen (O) Now, count the oxygen atoms on the product side: 1 CO₂ has 2 O atoms, and 2 H₂O have 2 O atoms, for a total of 4 O atoms. On the reactant side, we already have 1 O atom in CH₃OH. We need 3 more O atoms to balance. To get 3 O atoms from O₂, we need ³⁄₂ molecules of O₂. \[ \mathrm{CH_3OH} + \frac{3}{2}~\mathrm{O_2} \longrightarrow \mathrm{CO_2} + \mathrm{2~H_2O} \]

Step 4: Clear the Fraction The standard convention is to use the smallest whole-number coefficients. To eliminate the fraction, multiply the entire equation by 2. \[ 2 \times \left( \mathrm{CH_3OH} + \frac{3}{2}~\mathrm{O_2} \longrightarrow \mathrm{CO_2} + \mathrm{2~H_2O} \right) \] This gives the final, balanced equation: \[ \mathrm{2~CH_3OH} + \mathrm{3~O_2} \longrightarrow \mathrm{2~CO_2} + \mathrm{4~H_2O} \]

Write a balanced equation for the following reaction by placing appropriate stoichiometric coefficients.

\[\begin{align*} {\color{currentcolor}\rule[-1.0pt]{2em}{0.5pt}}~\mathrm{CH_3NHNH_2(l)} ~+~ {\color{currentcolor}\rule[-1.0pt]{2em}{0.5pt}}~\mathrm{O_2(g)} ~\longrightarrow~ {\color{currentcolor}\rule[-1.0pt]{2em}{0.5pt}}~\mathrm{CO_2(g)} ~+~ {\color{currentcolor}\rule[-1.0pt]{2em}{0.5pt}}~\mathrm{H_2O(g)} ~+~ {\color{currentcolor}\rule[-1.0pt]{2em}{0.5pt}}~\mathrm{N_2(g)} \end{align*}\]

Solution

Answer:

\[\begin{align*} \mathrm{2~CH_3NHNH_2(l)} + \mathrm{5~O_2(g)} \longrightarrow \mathrm{2~CO_2(g)} + \mathrm{6~H_2O(g)} + \mathrm{2~N_2(g)} \end{align*}\]

Concept: balancing equations

A reliable method for balancing the combustion of a compound containing C, H, and N is to balance the elements that appear in only one compound on each side first (C, H, N), and save the pure element (O₂) for last. We can show this step-by-step.

1. Balance C, H, and N:

   • Balance C: 1 C on left → 1 CO₂ on right.
   • Balance H: 6 H on left → 3 H₂O on right.
   • Balance N: 2 N on left → 1 N₂ on right.

   This gives the intermediate equation with the main elements balanced: \[ \mathrm{1~CH_3NHNH_2} + \mathrm{O_2} \longrightarrow \mathrm{1~CO_2} + \mathrm{3~H_2O} + \mathrm{1~N_2} \]

2. Balance O: Next, count the total oxygen atoms on the product side: (1 × 2 O in CO₂) + (3 × 1 O in H₂O) = 5 oxygen atoms. To get 5 oxygen atoms on the reactant side, we need ⁵⁄₂ molecules of O₂. \[ \mathrm{CH_3NHNH_2} + \frac{5}{2}~\mathrm{O_2} \longrightarrow \mathrm{CO_2} + \mathrm{3~H_2O} + \mathrm{N_2} \]

3. Clear the Fraction: The standard convention is to use the smallest whole-number coefficients. To eliminate the 5/2 fraction, multiply every coefficient in the entire equation by 2. \[ 2 \left( \mathrm{1~CH_3NHNH_2} + \frac{5}{2}~\mathrm{O_2} \longrightarrow \mathrm{1~CO_2} + \mathrm{3~H_2O} + \mathrm{1~N_2} \right) \] This yields the final, correctly balanced equation \[ \mathrm{2~CH_3NHNH_2(l)} + \mathrm{5~O_2(g)} \longrightarrow \mathrm{2~CO_2(g)} + \mathrm{6~H_2O(g)} + \mathrm{2~N_2(g)} \]

Write a balanced equation for the following reaction by placing appropriate stoichiometric coefficients.

\[\begin{align*} {\color{currentcolor}\rule[-1.0pt]{2em}{0.5pt}}~\mathrm{Se} ~+~ {\color{currentcolor}\rule[-1.0pt]{2em}{0.5pt}}~\mathrm{BrF_5} ~\longrightarrow~ {\color{currentcolor}\rule[-1.0pt]{2em}{0.5pt}}~\mathrm{SeF_6} ~+~ {\color{currentcolor}\rule[-1.0pt]{2em}{0.5pt}}~\mathrm{BrF_3} \end{align*}\]

Solution

Answer:

\[\begin{align*} \mathrm{1~Se} + \mathrm{3~BrF_5} \longrightarrow \mathrm{1~SeF_6} + \mathrm{3~BrF_3} \end{align*}\]

Concept: balancing equations

This equation can be difficult to balance by simple inspection. A more systematic approach is the algebraic method, where we assign variable coefficients to each species and solve a system of equations.

Step 1: Assign Variable Coefficients Let’s assign a variable to each species in the reaction: \[ a~\mathrm{Se} + b~\mathrm{BrF_5} \longrightarrow c~\mathrm{SeF_6} + d~\mathrm{BrF_3} \]

Step 2: Create Atom Balance Equations We write an equation for each element, ensuring the number of atoms is the same on both sides. * For Se: a = c * For Br: b = d * For F: 5b = 6c + 3d

Step 3: Solve the System of Equations Since we are looking for a ratio, we can start by setting one of the coefficients equal to 1. Let’s choose a = 1. * If a = 1, then from the Se balance, c = 1. * Now substitute c = 1 and b = d into the fluorine balance equation: \[ \begin{align*} 5b &= 6(1) + 3(b) \\[1.5ex] 5b &= 6 + 3b \\[1.5ex] 2b &= 6 \\[1.5ex] b &= 3 \end{align*} \] * Since b = d, then d = 3.

Step 4: Write the Final Balanced Equation We have found the coefficients: a = 1, b = 3, c = 1, and d = 3. We place these back into the equation. \[ \mathrm{1~Se} + \mathrm{3~BrF_5} \longrightarrow \mathrm{1~SeF_6} + \mathrm{3~BrF_3} \]

Determine the proper value for m and n needed to balance each equation.

\[ \begin{align*} \mathrm{B}_m\mathrm{H}_n + \mathrm{3~O_2} \longrightarrow \mathrm{B_2O_3} + \mathrm{3~H_2O} \end{align*} \]

Solution

Answer: m = 2; n = 6

Concept: balancing equations

To find the values of m and n, we must ensure that the number of atoms for each element is the same on both the reactant and product sides of the equation. We can solve for m and n by balancing the boron and hydrogen atoms independently.

Step 1: Balance the Boron (B) Atoms

• Product Side: There is one molecule of B₂O₃, which contains 2 boron atoms.
• Reactant Side: To balance this, the B_mH_n molecule must also provide 2 boron atoms. Since there is only one molecule of B_mH_n, the subscript m must be 2.
  • Therefore, m = 2.

Step 2: Balance the Hydrogen (H) Atoms

• Product Side: There are three molecules of H₂O. The total number of hydrogen atoms is 3 × 2 = 6 hydrogen atoms.
• Reactant Side: To balance this, the B₂H_n molecule must also provide 6 hydrogen atoms. Since there is only one molecule, the subscript n must be 6.
  • Therefore, n = 6.

Step 3: Write the Final Balanced Equation

By substituting m=2 and n=6, we get the complete formula for the reactant, diborane (B₂H₆), and the fully balanced equation. \[ \mathrm{B_2H_6} + \mathrm{3~O_2} \longrightarrow \mathrm{B_2O_3} + \mathrm{3~H_2O} \] (A final check of the oxygen atoms (6 on each side) confirms our answer is correct.)

Determine the proper value for m and n needed to balance each equation.

\[ \begin{align*} \mathrm{H}_m\mathrm{IO}_n ~\longrightarrow~ \mathrm{H^+} + \mathrm{IO_4^-} + \mathrm{2~H_2O} \end{align*} \]

Solution

Answer: m = 5; n = 6

Concept: balancing equations

To find the values of m and n, we must ensure that the number of atoms for each element is the same on both the reactant and product sides of the equation. We can solve for m and n by balancing the hydrogen and oxygen atoms independently.

Step 1: Balance the Hydrogen (H) Atoms

• Product Side: Count the total number of hydrogen atoms on the right side of the equation.
  • One H⁺ ion contributes 1 H atom.
  • Two H₂O molecules contribute 2 × 2 = 4 H atoms.
  • Total H on the right = 1 + 4 = 5 hydrogen atoms.
• Reactant Side: To balance this, the H_mIO_n molecule must also provide 5 hydrogen atoms. Therefore, the subscript m must be 5.
  • Therefore, m = 5.

Step 2: Balance the Oxygen (O) Atoms

• Product Side: Count the total number of oxygen atoms on the right side.
  • One IO₄^– ion contributes 4 O atoms.
  • Two H₂O molecules contribute 2 × 1 = 2 O atoms.
  • Total O on the right = 4 + 2 = 6 oxygen atoms.
• Reactant Side: To balance this, the H₅IO_n molecule must also provide 6 oxygen atoms. Therefore, the subscript n must be 6.
  • Therefore, n = 6.

Step 3: Write the Final Balanced Equation

By substituting m=5 and n=6, we get the complete formula for the reactant, periodic acid (H₅IO₆), and the fully balanced equation. \[ \mathrm{H_5IO_6} \longrightarrow \mathrm{H^+} + \mathrm{IO_4^-} + \mathrm{2~H_2O} \] (A final check of the iodine atoms (1 on each side) and the net charge (0 on each side) confirms our answer is correct.)

Write a balanced equation for the following reaction by placing appropriate stoichiometric coefficients.

\[ \begin{align*} {\color{currentcolor}\rule[-1.0pt]{2em}{0.5pt}}~\mathrm{CH_4} ~+~ {\color{currentcolor}\rule[-1.0pt]{2em}{0.5pt}}~\mathrm{H_2O} ~\longrightarrow~ {\color{currentcolor}\rule[-1.0pt]{2em}{0.5pt}}~\mathrm{CO} ~+~ {\color{currentcolor}\rule[-1.0pt]{2em}{0.5pt}}~\mathrm{H_2} \end{align*} \]

Solution

Answer:

\[ \begin{align*} \mathrm{1~CH_4} + \mathrm{1~H_2O} \longrightarrow \mathrm{1~CO} + \mathrm{3~H_2} \end{align*} \]

Concept: balancing equations

We can balance this equation by sequentially balancing each element.

Step 1: Balance Carbon (C) Start with the unbalanced equation. There is 1 C atom on the left (in CH₄) and 1 C on the right (in CO). The carbons are already balanced. \[ \mathrm{1~CH_4} + \mathrm{H_2O} \longrightarrow \mathrm{1~CO} + \mathrm{H_2} \]

Step 2: Balance Oxygen (O) There is 1 O atom on the left (in H₂O) and 1 O on the right (in CO). The oxygens are also already balanced. \[ \mathrm{1~CH_4} + \mathrm{1~H_2O} \longrightarrow \mathrm{1~CO} + \mathrm{H_2} \]

Step 3: Balance Hydrogen (H) Now, count the hydrogen atoms on the reactant side. There are 4 H (from CH₄) + 2 H (from H₂O) for a total of 6 H atoms. To get 6 H atoms on the product side, we need 3 molecules of H₂, since each contains 2 H atoms. \[ \mathrm{1~CH_4} + \mathrm{1~H_2O} \longrightarrow \mathrm{1~CO} + \mathrm{3~H_2} \]

Step 4: Final Check * Reactants: 1 C, 6 H, 1 O * Products: 1 C, 6 H, 1 O The equation is now fully balanced.

Write a balanced equation for the following reaction by placing appropriate stoichiometric coefficients.

\[ \begin{align*} {\color{currentcolor}\rule[-1.0pt]{2em}{0.5pt}}~\mathrm{Ag_2O(s)} ~\longrightarrow~ {\color{currentcolor}\rule[-1.0pt]{2em}{0.5pt}}~\mathrm{Ag(s)} ~+~ {\color{currentcolor}\rule[-1.0pt]{2em}{0.5pt}}~\mathrm{O_2} \end{align*} \]

Solution

Answer:

\[\begin{align*} \mathrm{2~Ag_2O(s)} \longrightarrow \mathrm{4~Ag(s)} + \mathrm{1~O_2} \end{align*}\]

Concept: balancing equations

We can balance this decomposition reaction by balancing the elements one at a time. A good strategy is often to balance the element that appears in the most compounds first.

Step 1: Balance Silver (Ag) Start with the unbalanced equation. There are 2 Ag atoms on the left (in Ag₂O) and only 1 Ag atom on the right. To balance the silver, place a coefficient of 2 in front of Ag(s). \[ \mathrm{Ag_2O} \longrightarrow \mathrm{2~Ag} + \mathrm{O_2} \]

Step 2: Balance Oxygen (O) Now, look at the oxygen atoms. There is 1 O atom on the left side, but 2 O atoms on the right (in O₂). To balance this, we must place a coefficient of 2 in front of the Ag₂O on the left side. \[ \mathrm{2~Ag_2O} \longrightarrow \mathrm{2~Ag} + \mathrm{O_2} \]

Step 3: Re-check Silver (Ag) By changing the coefficient for Ag₂O, we have now unbalanced the silver. The left side now has 2 × 2 = 4 Ag atoms, while the right side still has only 2. We must update the coefficient for Ag(s) to 4. \[ \mathrm{2~Ag_2O} \longrightarrow \mathrm{4~Ag} + \mathrm{O_2} \]

Step 4: Final Check * Reactants: 4 Ag, 2 O * Products: 4 Ag, 2 O The equation is now fully balanced.

Chemical equations must be balanced because the resulting coefficients allow us to predict (select all that apply).

1. the amount of product that can form from a given amount of reactant.
2. whether the reaction requires a catalyst or not
3. how much of one reactant is required to react with a given amount of another
4. how much reactants are required to form a given amount of products
5. whether the given reaction is possible or not

Solution

Answer: A, C, D

Concept: balancing equations

A balanced chemical equation is the cornerstone of stoichiometry, which is the quantitative study of the relationships between the amounts of reactants and products. The coefficients in a balanced equation provide the mole ratios, which are the conversion factors used to make all quantitative predictions about a reaction.

The correct statements are all direct applications of stoichiometry: * (a), (c), and (d) are all fundamentally the same concept. The balanced coefficients allow us to calculate the amount of any substance in the reaction (whether reactant or product) from the known amount of any other substance.

The incorrect statements relate to different fields of chemistry: * (b) Predicting the need for a catalyst relates to chemical kinetics, the study of reaction rates. A balanced equation provides no information about how fast a reaction will proceed. * (e) Predicting whether a reaction is possible (spontaneous) relates to thermodynamics, the study of energy and spontaneity. A reaction can be perfectly balanced on paper but may not occur under a given set of conditions.

Choose the best reaction equation that coincides with the following reaction.

aqueous silver sulfate + aqueous barium iodide → solid barium sulfate + solid silver iodide

1. Ag₂SO₄ + BaI₂ → BaSO₄ + AgI
2. Ag₂SO₄(l) + BaI₂(l) → BaSO₄(s) + 2 AgI(s)
3. Ag₂SO₄(aq) + BaI₂(aq) → BaSO₄(s) + 2 AgI(s)
4. AgSO₄(aq) + BaI(aq) → BaSO₄(s) + 2 AgI(s)
5. AgSO₄(l) + 2 BaI(l) → Ba₂SO₄(s) + 2 AgI₂(s)

Solution

Answer: C

Concept: balancing equations; nomenclature

To solve this problem, we must perform three sequential steps: 1. Write the correct chemical formula for each named compound. 2. Assign the correct state symbol (aq) or (s) to each compound as described. 3. Balance the final equation by ensuring the number of atoms of each element is the same on both sides.

Step 1: Write the Chemical Formulas

• Silver sulfate: Silver is a transition metal that almost always forms a +1 ion (Ag⁺). Sulfate is the polyatomic anion SO₄^2–. To balance the charge, two Ag⁺ ions are needed. Formula: Ag₂SO₄.
• Barium iodide: Barium (Group 2) forms a Ba²⁺ ion. Iodide (Group 17) forms an I^– ion. Two I^– ions are needed. Formula: BaI₂.
• Barium sulfate: Ba²⁺ and SO₄^2–. The charges balance 1:1. Formula: BaSO₄.
• Silver iodide: Ag⁺ and I^–. The charges balance 1:1. Formula: AgI.

Step 2: Write the Unbalanced Equation with State Symbols

Using the formulas from Step 1 and the states given in the problem: \[ \mathrm{Ag_2SO_4(aq)} + \mathrm{BaI_2(aq)} \longrightarrow \mathrm{BaSO_4(s)} + \mathrm{AgI(s)} \]

Step 3: Balance the Equation

We inspect the atom count on both sides.

• Reactants: 2 Ag, 1 S, 4 O, 1 Ba, 2 I
• Products: 1 Ba, 1 S, 4 O, 1 Ag, 1 I

The silver (Ag) and iodine (I) atoms are unbalanced. To balance the 2 Ag and 2 I on the left side, we must place a coefficient of 2 in front of AgI on the right side. \[ \mathrm{Ag_2SO_4(aq)} + \mathrm{BaI_2(aq)} \longrightarrow \mathrm{BaSO_4(s)} + 2~\mathrm{AgI(s)} \] A final check shows all atoms are now balanced. This matches answer choice (c).

Chemical Reactions and Solubility

Classify each of the following (all that apply) as a

 I. strong electrolyte
 II. weak electrolyte
 III. nonelectrolyte
 IV. strong acid
 V. strong base
 VI. weak acid
 VII. weak base
 VIII. ionic compound
 IX. organic compound

1. HBr
2. ammonium carbonate
3. NaClO₄
4. ethanol
5. acetic acid
6. NH₃

Solution

Answer:
a. I, IV
b. I, VIII
c. I, VIII
d. III, IX
e. II, VI
f. II, VII

Concept: electrolytes; solubility rules

This problem requires applying several classification schemes based on how a substance behaves when dissolved in water.

• a. HBr (Hydrobromic acid): It is one of the seven common strong acids (IV). Because it dissociates 100% in water to form ions (H⁺ and Br⁻), it is a strong electrolyte (I).

• b. Ammonium carbonate ((NH₄)₂CO₃): It is a salt composed of the ammonium cation (NH₄⁺) and the carbonate anion (CO₃^2–), making it an ionic compound (VIII). According to the solubility rules, it is soluble in water. Therefore, it dissociates completely and is a strong electrolyte (I).

• c. NaClO₄ (Sodium perchlorate): It is a soluble salt composed of a metal cation (Na⁺) and a polyatomic anion (ClO₄^–), making it an ionic compound (VIII) and a strong electrolyte (I).

• d. Ethanol (C₂H₅OH): It is a compound primarily of carbon and hydrogen, making it an organic compound (IX). It dissolves in water but does not produce ions. Therefore, it is a nonelectrolyte (III).

• e. Acetic acid (CH₃COOH): It is a weak acid (VI). By definition, weak acids only partially ionize in water, making them weak electrolytes (II).

• f. NH₃ (Ammonia): It is the most common example of a weak base (VII). It reacts with water to a small extent to form ions (NH₄⁺ and OH^–), making it a weak electrolyte (II).

Which of the following compounds are insoluble in water? Select all that apply.

1. CoCO₃
2. Cu₃(PO₄)₂
3. AgNO₃
4. Na₂S
5. AgI

Solution

Answer: A, B, E

Concept: solubility rules

To determine solubility, we apply the general solubility rules for ionic compounds in water.

Insoluble Compounds:

• (a) CoCO₃: This compound is insoluble. The rule is that most carbonate (CO₃²⁻) salts are insoluble. Cobalt is not one of the exceptions (Group 1 cations, NH₄⁺).
• (b) Cu₃(PO₄)₂: This compound is insoluble. The rule is that most phosphate (PO₄³⁻) salts are insoluble. Copper is not one of the exceptions.
• (e) AgI: This compound is insoluble. While most iodide (I⁻) salts are soluble, silver (Ag⁺) is one of the key exceptions to this rule.

Soluble Compounds:

• (c) AgNO₃: This compound is soluble. The rule is that all nitrate (NO₃⁻) salts are soluble, without exception.
• (d) Na₂S: This compound is soluble. The rule is that all salts containing an alkali metal (Group 1) cation, like Na⁺, are soluble.

Which of the following combinations will form a precipitate? Select all that apply.

1. SrCl₂(aq) + Na₂S(aq)
2. KCl(aq) + CaS(aq)
3. Hg(NO₃)₂(aq) + Na₃PO₄(aq)
4. Ba(NO₃)₂(aq) + KOH(aq)
5. NaOH(aq) + FeCl₃(aq)

Solution

Answer: C and E

Concept: solubility rules; precipitation

To solve this, we must first determine the two potential products of each double displacement reaction by swapping the cations and anions. Then, we use the solubility rules to check if either of the new compounds is insoluble (a precipitate).

---

a. SrCl₂(aq) + Na₂S(aq) → SrS(aq) + 2 NaCl(aq)

• Potential Products: SrS and NaCl.
• Solubility Check:
  • NaCl: Soluble (Rule: Group 1 cations are always soluble).
  • SrS: Soluble (Rule: Sulfides of Group 2 metals are an exception to the general insolubility of sulfides).
• Conclusion: No precipitate forms.

---

b. 2 KCl(aq) + CaS(aq) → K₂S(aq) + CaCl₂(aq)

• Potential Products: K₂S and CaCl₂.
• Solubility Check:
  • K₂S: Soluble (Rule: Group 1 cations).
  • CaCl₂: Soluble (Rule: Chlorides are soluble, and Ca²⁺ is not an exception).
• Conclusion: No precipitate forms.

---

c. 3 Hg(NO₃)₂(aq) + 2 Na₃PO₄(aq) → Hg₃(PO₄)₂(s) + 6 NaNO₃(aq)

• Potential Products: Hg₃(PO₄)₂ and NaNO₃.
• Solubility Check:
  • NaNO₃: Soluble (Rule: Group 1 cations and nitrates are always soluble).
  • Hg₃(PO₄)₂: Insoluble (Rule: Phosphates are generally insoluble).
• Conclusion: A precipitate of mercury(II) phosphate forms.

---

d. Ba(NO₃)₂(aq) + 2 KOH(aq) → Ba(OH)₂(aq) + 2 KNO₃(aq)

• Potential Products: Ba(OH)₂ and KNO₃.
• Solubility Check:
  • KNO₃: Soluble (Rule: Group 1 and nitrates).
  • Ba(OH)₂: Soluble (Rule: Hydroxides are generally insoluble, but Ba²⁺ is a key exception).
• Conclusion: No precipitate forms.

---

e. 3 NaOH(aq) + FeCl₃(aq) → Fe(OH)₃(s) + 3 NaCl(aq)

• Potential Products: Fe(OH)₃ and NaCl.
• Solubility Check:
  • NaCl: Soluble (Rule: Group 1).
  • Fe(OH)₃: Insoluble (Rule: Hydroxides are generally insoluble, and Fe³⁺ is not an exception).
• Conclusion: A precipitate of iron(III) hydroxide forms.

Calcium chloride is used to “salt” streets in the winter to melt ice and snow. Write a net ionic reaction to show how this substance breaks apart when it dissolves in water.

Solution

Answer:

The process of an ionic solid breaking apart into its constituent ions as it dissolves in a solvent is called dissociation. According to solubility rules, calcium chloride (CaCl₂) is a soluble salt.

When solid CaCl₂ is added to water, it dissociates into its ions: one calcium cation (Ca²⁺) and two chloride anions (Cl⁻). It is important to include the state symbols to show the physical state of the species before and after dissolving.

The balanced dissociation equation is:

\[ \begin{align*} \mathrm{CaCl_2(s)} \longrightarrow \mathrm{Ca^{2+}(aq)} + \mathrm{2~Cl^-(aq)} \end{align*} \]

Concept: net ionic equation

Lead(II) nitrate reacts with sodium chloride. Choose the net ionic equation for the reaction.

1. Pb²⁺(aq) + Cl^–(aq) → PbCl(s)
2. Pb²⁺(aq) + NO₃^–(aq) + Na⁺(aq) + Cl^–(aq) → PbCl(s) + NaNO₃(aq)
3. PbNO₃(aq) + NaCl(aq) → PbCl(s) + NaNO₃(aq)
4. Pb²⁺(aq) + 2 Cl^–(aq) → PbCl₂(s)

Solution

Answer: D

Concept: solubility rules; precipitation

Molecular equation:

\[ \mathrm{Pb(NO_3)_2(aq)} + 2~\mathrm{NaCl(aq)} \longrightarrow \mathrm{PbCl_2(s)} + 2~\mathrm{NaNO_3(aq)} \]

Complete ionic equation:

\[ \mathrm{Pb^{2+}(aq) + 2~NO_3{^-}(aq) + 2~Na^+(aq) + 2~Cl^-(aq) \longrightarrow PbCl_2(s) + 2~Na^+(aq) + 2~NO_3{^-}(aq)} \]

Net ionic equation:

\[ \mathrm{Pb^{2+}(aq) + 2~Cl^-(aq) \longrightarrow PbCl_2(s)} \]

Aqueous solutions of sodium sulfate and barium chloride react. What is the sum of the coefficients from the balanced net ionic equation?

Solution

Answer: 3

Concept: solubility rules; precipitation; net ionic equation; balancing equations

Molecular equation:

\[ \mathrm{Na_2SO_4(aq)} + \mathrm{BaCl_2(aq)} \longrightarrow 2~\mathrm{NaCl(aq)} + \mathrm{BaSO_4(s)} \]

Complete ionic equation:

\[ \mathrm{2~Na^+(aq)} + \mathrm{SO_4{^{2-}}(aq)} + \mathrm{Ba^{2+}(aq)} + \mathrm{2~Cl^-(aq)} \longrightarrow 2~\mathrm{Na^+(aq)} + 2~\mathrm{Cl^-(aq)} + \mathrm{BaSO_4(s)} \]

Net ionic equation:

\[ \mathrm{Ba^{2+}(aq)} + \mathrm{SO_4{^{2-}}(aq)} \longrightarrow \mathrm{BaSO_4(s)} \]

Sum the coefficients from the net ionic equation:

\[ 1 + 1 + 1 = 3 \]

Answer the questions for the following reaction.

\[ \mathrm{HI(aq)} + \mathrm{Ca(OH)_2(aq)} \longrightarrow \mathrm{H_2O(l)} + \mathrm{CaI_2(aq)} \]

1. Is the acid strong or weak?
2. Is the base strong or weak?
3. What is the net ionic equation for the reaction?

Solution

Answer:

1. strong
2. strong
3. H⁺(aq) + OH^–(aq) → H₂O(l)

Concept: solubility rules; precipitation; net ionic equation; balancing equations

Molecular equation:

\[ \mathrm{2~HI(aq)} + \mathrm{Ca(OH)_2(aq)} \longrightarrow \mathrm{2~H_2O(l)} + \mathrm{CaI_2(aq)} \]

Complete ionic equation:

\[ \mathrm{2~H^+(aq)} + \mathrm{2~I^-(aq)} + \mathrm{Ca^{2+}(aq)} + \mathrm{2~OH^-(aq)} \longrightarrow \mathrm{2~H_2O(l)} + \mathrm{Ca^{2+}(aq)} + \mathrm{2~I^-(aq)} \]

Net ionic equation:

\[ \mathrm{H^+(aq)} + \mathrm{OH^-(aq)} \longrightarrow \mathrm{H_2O(l)} \]

A reaction between hydrobromic acid and potassium hydroxide occurs.

1. Is the acid strong or weak?
2. Is the base strong or weak?
3. What is the net ionic equation for the reaction?

Solution

Answer:

1. strong
2. strong
3. H⁺(aq) + OH^–(aq) → H₂O(l)

Concept: acid-base reaction; net ionic equation; balancing equations

Molecular equation:

\[ \mathrm{HBr(aq)} + \mathrm{KOH(aq)} \longrightarrow \mathrm{KBr(aq)} + \mathrm{H_2O(l)} \]

Complete ionic equation:

\[ \mathrm{H^+(aq)} + \mathrm{Br^-(aq)} + \mathrm{K^+(aq)} + \mathrm{OH^-(aq)} \longrightarrow \mathrm{K^+(aq)} + \mathrm{Br^-(aq)} + \mathrm{H_2O(l)} \]

Net ionic equation:

\[ \mathrm{H^+(aq)} + \mathrm{OH^-(aq)} \longrightarrow \mathrm{H_2O(l)} \]

Which is the spectator ion in the reaction between potassium carbonate and calcium iodide? Select all that apply.

1. K⁺(aq)
2. CO₃^2–(aq)
3. Ca²⁺(aq)
4. I^–(aq)

Solution

Answer: A, D

Concept: solubility rules; precipitation; net ionic equation; balancing equations

Molecular equation:

\[ \mathrm{K_2CO_3(aq)} + \mathrm{CaI_2(aq)} \longrightarrow 2~\mathrm{KI(aq)} + \mathrm{CaCO_3(s)} \]

Complete ionic equation:

\[ \mathrm{2~K^+(aq)} + \mathrm{CO_3{^{2-}}(aq)} + \mathrm{Ca^{2+}(aq)} + \mathrm{2~I^-(aq)} \longrightarrow \mathrm{2~K^+(aq)} + 2~\mathrm{I^-(aq)} + \mathrm{CaCO_3(s)} \]

Net ionic equation:

\[ \mathrm{CO_3{^{2-}}(aq)} + \mathrm{Ca^{2+}(aq)} \longrightarrow \mathrm{CaCO_3(s)} \]

What is the sum of the coefficients of the net ionic equation for aqueous sodium hydroxide neutralized by aqueous acetic acid?

Solution

Answer: 4

Concept: acid-base reaction; net ionic equation; balancing equations

Molecular equation:

\[ \mathrm{CH_3COOH(aq)} + \mathrm{NaOH(aq)} \longrightarrow \mathrm{NaCH_3COO(aq)} + \mathrm{H_2O(l)} \]

Complete ionic equation:

\[ \mathrm{CH_3COOH(aq)} + \mathrm{Na^+(aq)} + \mathrm{OH^-(aq)} \longrightarrow \mathrm{Na^+(aq)} + \mathrm{CH_3COO^-(aq)} + \mathrm{H_2O(l)} \]

Net ionic equation:

\[ \mathrm{CH_3COOH(aq)} + \mathrm{OH^-(aq)} \longrightarrow \mathrm{CH_3COO^-(aq)} + \mathrm{H_2O(l)} \]

Sum the coefficients in the net ionic reaction.

\[ 1 + 1 + 1 + 1 = 4 \]

When the following solutions are mixed together, what precipitate (if any) will form?

1. FeSO₄(aq) + KCl(aq)
2. Al(NO₃)₃(aq) + Ba(OH)₂(aq)
3. CaCl₂(aq) + Na₂SO₄(aq)
4. K₂S(aq) + Ni(NO₃)₂(aq)
5. Hg₂(NO₃)₂(aq) + CuSO₄(aq)
6. Ni(NO₃)₂(aq) + CaCl₂(aq)
7. K₂CO₃(aq) + MgI₂(aq)
8. Na₂CrO₄(aq) + AlBr₃(aq)

Solution

Answer:

1. none
2. Al(OH)₃(s)
3. CaSO₄(s)
4. NiS(s)
5. Hg₂SO₄(s)
6. none
7. MgCO₃(s)
8. Al₂(CrO₄)₃(s)

Concept: solubility rules; double-displacement reaction

For each reaction, we first determine the potential products of the double displacement reaction and then use the solubility rules to identify any insoluble compounds (precipitates).

1. FeSO₄(aq) + 2 KCl(aq) → FeCl₂(aq) + K₂SO₄(aq)
   • Products: FeCl₂ and K₂SO₄.
   • Analysis: Both are soluble (chlorides are generally soluble; Group 1 salts are soluble).
   • Result: No precipitate.
2. 2 Al(NO₃)₃(aq) + 3 Ba(OH)₂(aq) → 2 Al(OH)₃(s) + 3 Ba(NO₃)₂(aq)
   • Products: Al(OH)₃ and Ba(NO₃)₂.
   • Analysis: Ba(NO₃)₂ is soluble (nitrates are soluble). Al(OH)₃ is insoluble (most hydroxides are insoluble).
   • Result: Precipitate is Al(OH)₃.
3. CaCl₂(aq) + Na₂SO₄(aq) → CaSO₄(s) + 2 NaCl(aq)
   • Products: CaSO₄ and NaCl.
   • Analysis: NaCl is soluble (Group 1). CaSO₄ is insoluble (sulfates are generally soluble, but Ca²⁺ is a key exception).
   • Result: Precipitate is CaSO₄.
4. K₂S(aq) + Ni(NO₃)₂(aq) → 2 KNO₃(aq) + NiS(s)
   • Products: KNO₃ and NiS.
   • Analysis: KNO₃ is soluble (Group 1 and nitrates). NiS is insoluble (most sulfides are insoluble).
   • Result: Precipitate is NiS.
5. Hg₂(NO₃)₂(aq) + CuSO₄(aq) → Hg₂SO₄(s) + Cu(NO₃)₂(aq)
   • Products: Hg₂SO₄ and Cu(NO₃)₂.
   • Analysis: Cu(NO₃)₂ is soluble (nitrates). Hg₂SO₄ is insoluble (sulfates are generally soluble, but the mercury(I) ion, Hg₂²⁺, is a key exception).
   • Result: Precipitate is Hg₂SO₄.
6. Ni(NO₃)₂(aq) + CaCl₂(aq) → NiCl₂(aq) + Ca(NO₃)₂(aq)
   • Products: NiCl₂ and Ca(NO₃)₂.
   • Analysis: Both are soluble (nitrates are soluble; chlorides are soluble).
   • Result: No precipitate.
7. K₂CO₃(aq) + MgI₂(aq) → 2 KI(aq) + MgCO₃(s)
   • Products: KI and MgCO₃.
   • Analysis: KI is soluble (Group 1). MgCO₃ is insoluble (most carbonates are insoluble).
   • Result: Precipitate is MgCO₃.
8. 3 Na₂CrO₄(aq) + 2 AlBr₃(aq) → 6 NaBr(aq) + Al₂(CrO₄)₃(s)
   • Products: NaBr and Al₂(CrO₄)₃.
   • Analysis: NaBr is soluble (Group 1). Al₂(CrO₄)₃ is insoluble (most chromates are insoluble).
   • Result: Precipitate is Al₂(CrO₄)₃.

Which of the following substances are soluble in water? Select all that apply.

1. aluminum nitrate
2. magnesium chloride
3. rubidium sulfate
4. nickel(II) hydroxide
5. lead(II) sulfide
6. barium hydroxide
7. iron(III) phosphate

Solution

Answer: A, B, C, F

Concept: solubility rules

To solve this, we must first write the correct chemical formula from the name and then apply the general solubility rules.

Soluble Compounds:

• (a) Aluminum nitrate (Al(NO₃)₃): This is soluble. Rule: All nitrate (NO₃^–) salts are soluble, without exception.
• (b) Magnesium chloride (MgCl₂): This is soluble. Rule: Most chloride (Cl^–) salts are soluble. Mg²⁺ is not one of the exceptions.
• (c) Rubidium sulfate (Rb₂SO₄): This is soluble. Rule: All salts containing an alkali metal (Group 1) cation, like Rb⁺, are soluble.
• (f) Barium hydroxide (Ba(OH)₂): This is soluble. Rule: While most hydroxide (OH^–) salts are insoluble, the hydroxides of the heavy Group 2 metals (Ca²⁺, Sr²⁺, and Ba²⁺) are important exceptions.

Insoluble Compounds:

• (d) Nickel(II) hydroxide (Ni(OH)₂): This is insoluble. Rule: Most hydroxide salts are insoluble. Ni²⁺ is not an exception.
• (e) Lead(II) sulfide (PbS): This is insoluble. Rule: Most sulfide (S^2–) salts are insoluble. Pb²⁺ is not one of the exceptions.
• (g) Iron(III) phosphate (FePO₄): This is insoluble. Rule: Most phosphate (PO₄^3–) salts are insoluble. Fe³⁺ is not an exception.

Write the net ionic equations for the following reactions:

1. ammonium sulfate and barium nitrate
2. lead(II) nitrate and sodium chloride
3. sodium phosphate and potassium nitrate
4. sodium bromide and rubidium chloride
5. copper(II) chloride and sodium hydroxide

Solution

Answer:

1. Ba²⁺(aq) + SO₄^2–(aq) → BaSO₄(s)
2. Pb²⁺(aq) + 2 Cl^–(aq) → PbCl₂(s)
3. none (all ions are spectators)
4. none (all ions are spectators)
5. Cu²⁺(aq) + 2 OH^–(aq) → Cu(OH)₂(s)

Concept: net ionic equation; balancing equations; solubility rules

Write the balanced molecular equation, complete ionic equation, and net ionic equation for the following acid-base reactions.

1. HClO₄(aq) + Mg(OH)₂(s)
2. HCN(aq) + NaOH(aq)
3. HCl(aq) + NaOH(aq)

Solution

Answer:

A

Molecular equation

\[ \begin{align*} 2~\mathrm{HClO_4(aq)} + \mathrm{Mg(OH)_2(s)} &\longrightarrow 2~\mathrm{H_2O(l)} + \mathrm{Mg(ClO_4)_2(aq)} \end{align*} \]

Complete ionic equation

\[ \begin{align*} \mathrm{2~H^+(aq)} + \mathrm{2~ClO_4{^-}(aq)} + \mathrm{Mg(OH)_2(s)} &\longrightarrow \mathrm{2~H_2O(l)} + \mathrm{Mg^{2+}(aq)} + 2~\mathrm{ClO_4{^-}(aq)} \end{align*} \]

Net ionic equation

\[ \begin{align*} \mathrm{2~H^{+}(aq)} + \mathrm{Mg(OH)_2(s)} &\longrightarrow \mathrm{2~H_2O(l)} + \mathrm{Mg^{2+}(aq)} \end{align*} \]

B

Molecular equation

\[ \begin{align*} \mathrm{HCN(aq)} + \mathrm{NaOH(aq)} &\longrightarrow \mathrm{H_2O(l)} + \mathrm{NaCN(aq)} \end{align*} \] Complete ionic equation

\[ \begin{align*} \mathrm{HCN(aq)} + \mathrm{Na^+(aq)} + \mathrm{OH^-(aq)} &\longrightarrow \mathrm{H_2O(l)} + \mathrm{Na^{+}(aq)} + \mathrm{CN^-(aq)} \end{align*} \]

Net ionic equation

\[ \begin{align*} \mathrm{HCN(aq)} + \mathrm{OH^-(aq)} &\longrightarrow \mathrm{H_2O(l)} + \mathrm{CN^-(aq)} \end{align*} \]

C

Molecular equation

\[ \begin{align*} \mathrm{HCl(aq)} + \mathrm{NaOH(aq)} &\longrightarrow \mathrm{H_2O(l)} + \mathrm{NaCl(aq)} \end{align*} \]

Complete ionic equation

\[ \begin{align*} \mathrm{H^+(aq)} + \mathrm{Cl^-(aq)} + \mathrm{Na^+(aq)} + \mathrm{OH^-(aq)} &\longrightarrow \mathrm{H_2O(l)} + \mathrm{Na^{+}(aq)} + \mathrm{Cl^-(aq)} \end{align*} \]

Net ionic equation

\[ \begin{align*} \mathrm{H^{+}(aq)} + \mathrm{OH^-(aq)} &\longrightarrow \mathrm{H_2O(l)} \end{align*} \]

Concept: acid-base reaction; net ionic equation

Write a balanced chemical equation between an acid and a base that would have the following salt appear as a product.

1. potassium perchlorate
2. cesium nitrate
3. calcium iodide

Solution

Answer:

A

\[ \begin{align*} \mathrm{HClO_4(aq)} + \mathrm{KOH(aq)} &\longrightarrow \mathrm{H_2O(l)} + \mathrm{KClO_4(aq)} \end{align*} \]

B

\[ \begin{align*} \mathrm{HNO_3(aq)} + \mathrm{CsOH(aq)} &\longrightarrow \mathrm{H_2O(l)} + \mathrm{CsNO_3(aq)} \end{align*} \]

C

\[ \begin{align*} \mathrm{2~HI(aq)} + \mathrm{Ca(OH)_2(aq)} &\longrightarrow \mathrm{2~H_2O(l)} + \mathrm{CaI_2(aq)} \end{align*} \]

Concept: acid-base reaction; salts

Oxidation-reduction

Assign oxidation states for all atoms in each of the following compounds.

1. KMnO₄
2. NiO₂
3. Na₄Fe(OH)₆
4. (NH₄)₂HPO₄
5. P₄O₆

Solution

Answer:

The first row of numbers represents the oxidation state for each atom.

A

\[ \begin{align*} &\underset{\begin{subarray}{c} +1 \\ +1 \end{subarray}}{\phantom{_1}\mathrm{K}\phantom{_1}} \underset{\begin{subarray}{c} +7 \\ +7 \end{subarray}}{\phantom{_1}\mathrm{Mn}\phantom{_1}} \underset{\begin{subarray}{c} -2 \\ -8 \end{subarray}}{\phantom{_1}\mathrm{O_4}\phantom{_1}} \end{align*} \]

B

\[ \begin{align*} &\underset{\begin{subarray}{c} +4 \\ +4 \end{subarray}}{\phantom{_1}\mathrm{Ni}\phantom{_1}} \underset{\begin{subarray}{c} -2 \\ -4 \end{subarray}}{\phantom{_1}\mathrm{O_2}\phantom{_1}} \end{align*} \]

C

\[ \begin{align*} &\underset{\begin{subarray}{c} +1 \\ +4 \end{subarray}}{\phantom{_1}\mathrm{Na_4}\phantom{_1}} \underset{\begin{subarray}{c} +2 \\ +2 \end{subarray}}{\phantom{_1}\mathrm{Fe}\phantom{_1}} \underset{\begin{subarray}{c} -2 \\ -12 \end{subarray}}{(\phantom{_1}\mathrm{O}\phantom{_1}} \underset{\begin{subarray}{c} +1 \\ +6 \end{subarray}}{\phantom{_1}\mathrm{H}\phantom{_1})_6} \end{align*} \]

D

\[ \begin{align*} &\underset{\begin{subarray}{c} -3 \\ -3 \end{subarray}}{( \phantom{_1}\mathrm{N}\phantom{_1}} \underset{\begin{subarray}{c} +1 \\ +4 \end{subarray}}{\phantom{_1}\mathrm{H_4}\phantom{_1}})_2 \underset{\begin{subarray}{c} +1 \\ +1 \end{subarray}}{\phantom{_1}\mathrm{H}\phantom{_1}} \underset{\begin{subarray}{c} +5 \\ +5 \end{subarray}}{\phantom{_1}\mathrm{P}\phantom{_1}} \underset{\begin{subarray}{c} -2 \\ -8 \end{subarray}}{\phantom{_1}\mathrm{O_4}\phantom{_1}} \end{align*} \]

E

\[ \begin{align*} &\underset{\begin{subarray}{c} +3 \\ +12 \end{subarray}}{\phantom{_1}\mathrm{P_4}\phantom{_1}} \underset{\begin{subarray}{c} -2 \\ -12 \end{subarray}}{\phantom{_1}\mathrm{O_6}\phantom{_1}} \end{align*} \]

Concept: oxidation state

Assign oxidation states for all atoms in each of the following compounds.

1. Fe₃O₄
2. XeOF₄
3. SF₄
4. CO
5. C₆H₁₂O₆

Solution

Answer:

The first row of numbers represents the oxidation state for each atom.

A

\[ \begin{align*} &\underset{\begin{subarray}{c} +2/+3 \\ +8 \end{subarray}}{\phantom{_1}\mathrm{Fe_3}\phantom{_1}} \underset{\begin{subarray}{c} -2 \\ -8 \end{subarray}}{\phantom{_1}\mathrm{O_4}\phantom{_1}} \end{align*} \]

B

\[ \begin{align*} &\underset{\begin{subarray}{c} +6 \\ +6 \end{subarray}}{\phantom{_1}\mathrm{Xe}\phantom{_1}} \underset{\begin{subarray}{c} -2 \\ -2 \end{subarray}}{\phantom{_1}\mathrm{O}\phantom{_1}} \underset{\begin{subarray}{c} -1 \\ -4 \end{subarray}}{\phantom{_1}\mathrm{F_4}\phantom{_1}} \end{align*} \]

C

\[ \begin{align*} &\underset{\begin{subarray}{c} +4 \\ +4 \end{subarray}}{\phantom{_1}\mathrm{S}\phantom{_1}} \underset{\begin{subarray}{c} -1 \\ -4 \end{subarray}}{\phantom{_1}\mathrm{F_4}\phantom{_1}} \end{align*} \]

D

\[ \begin{align*} &\underset{\begin{subarray}{c} +2 \\ +2 \end{subarray}}{\phantom{_1}\mathrm{C}\phantom{_1}} \underset{\begin{subarray}{c} -2 \\ -2 \end{subarray}}{\phantom{_1}\mathrm{O}\phantom{_1}} \end{align*} \]

E

\[ \begin{align*} &\underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1}\mathrm{C_6}\phantom{_1}} \underset{\begin{subarray}{c} +1 \\ +12 \end{subarray}}{\phantom{_1}\mathrm{H_{12}}\phantom{_1}} \underset{\begin{subarray}{c} -2 \\ -12 \end{subarray}}{\phantom{_1}\mathrm{O_6}\phantom{_1}} \end{align*} \]

Concept: oxidation state

Specify which reactions are redox reactions and identify the oxidizing agent, reducing agent, the substance being oxidized, and the substance being reduced.

1. Cu(s) + 2 Ag⁺(aq) → 2 Ag(s) + Cu²⁺(aq)
2. HCl(g) + NH₃(g) → NH₄Cl(s)
3. SiCl₄(l) + 2 H₂O(l) → 4 HCl(aq) + SiO₂(s)
4. SiCl₄(l) + 2 Mg(s) → 2 MgCl₂(s) + Si(s)
5. Al(OH)₄^–(aq) → AlO₂^–(aq) + 2 H₂O(l)

Solution

Answer:

A

\[ \begin{align*} \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{Cu}\phantom{_1\!}}\mathrm{(s)} ~+~ \underset{\begin{subarray}{c} +1 \\ +1 \end{subarray}}{\phantom{_1\!}\mathrm{2~Ag^+}\phantom{_1\!}}\mathrm{(aq)} \longrightarrow \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{2~Ag}\phantom{_1\!}}\mathrm{(s)} ~+~ \underset{\begin{subarray}{c} +2 \\ +2 \end{subarray}}{\phantom{_1\!}\mathrm{Cu^{2+}}\phantom{_1\!}}\mathrm{(aq)} \end{align*} \]

Oxidizing agent: Ag⁺(aq); Reducing agent: Cu(s)
Substance oxidized: Cu(s); Substance reduced: Ag⁺(aq)

B

\[ \begin{align*} \underset{\begin{subarray}{c} +1 \\ +1 \end{subarray}}{\phantom{_1\!}\mathrm{H}\phantom{_1\!}} \underset{\begin{subarray}{c} -1 \\ -1 \end{subarray}}{\phantom{_1\!}\mathrm{Cl}\phantom{_1\!}}\mathrm{(g)} ~+~ \underset{\begin{subarray}{c} -3 \\ -3 \end{subarray}}{\phantom{_1\!}\mathrm{N}\phantom{_1\!}} \underset{\begin{subarray}{c} +1 \\ +3 \end{subarray}}{\phantom{_1\!}\mathrm{H_3}\phantom{_1\!}}\mathrm{(g)} \longrightarrow \underset{\begin{subarray}{c} -3 \\ -3 \end{subarray}}{\phantom{_1\!}\mathrm{N}\phantom{_1\!}} \underset{\begin{subarray}{c} +1 \\ +4 \end{subarray}}{\phantom{_1\!}\mathrm{H_4}\phantom{_1\!}} \underset{\begin{subarray}{c} -1 \\ -1 \end{subarray}}{\phantom{_1\!}\mathrm{Cl}\phantom{_1\!}}\mathrm{(s)} \end{align*} \]

Not a redox reaction.

C

\[ \begin{align*} \underset{\begin{subarray}{c} +4 \\ +4 \end{subarray}}{\phantom{_1\!}\mathrm{Si}\phantom{_1\!}} \underset{\begin{subarray}{c} -1 \\ -4 \end{subarray}}{\phantom{_1\!}\mathrm{Cl_4}\phantom{_1\!}}\mathrm{(l)} ~+~ \underset{\begin{subarray}{c} +1 \\ +2 \end{subarray}}{2~\phantom{_1\!}\mathrm{H_2}\phantom{_1\!}} \underset{\begin{subarray}{c} -2 \\ -2 \end{subarray}}{\phantom{_1\!}\mathrm{O}\phantom{_1\!}}\mathrm{(l)} \longrightarrow \underset{\begin{subarray}{c} +1 \\ +1 \end{subarray}}{4~\phantom{_1\!}\mathrm{H}\phantom{_1\!}} \underset{\begin{subarray}{c} -1 \\ -1 \end{subarray}}{\phantom{_1\!}\mathrm{Cl}\phantom{_1\!}}\mathrm{(aq)} ~+~ \underset{\begin{subarray}{c} +4 \\ +4 \end{subarray}}{\phantom{_1\!}\mathrm{Si}\phantom{_1\!}} \underset{\begin{subarray}{c} -2 \\ -4 \end{subarray}}{\phantom{_1\!}\mathrm{O_2}\phantom{_1\!}}\mathrm{(s)} \end{align*} \]

Not a redox reaction.

D

\[ \begin{align*} \underset{\begin{subarray}{c} +4 \\ +4 \end{subarray}}{\phantom{_1\!}\mathrm{Si}\phantom{_1\!}} \underset{\begin{subarray}{c} -1 \\ -4 \end{subarray}}{\phantom{_1\!}\mathrm{Cl_4}\phantom{_1\!}}\mathrm{(l)} ~+~ \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{2~\phantom{_1\!}\mathrm{Mg}\phantom{_1\!}}\mathrm{(s)} \longrightarrow \underset{\begin{subarray}{c} +2 \\ +2 \end{subarray}}{2~\phantom{_1\!}\mathrm{Mg}\phantom{_1\!}} \underset{\begin{subarray}{c} -1 \\ -2 \end{subarray}}{\phantom{_1\!}\mathrm{Cl_2}\phantom{_1\!}}\mathrm{(s)} ~+~ \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{Si}\phantom{_1\!}}\mathrm{(s)} \end{align*} \]

Oxidizing agent: SiCl₄(l); Reducing agent: Mg(s)
Substance oxidized: Mg(s); Substance reduced: SiCl₄(l)

E

\[ \begin{align*} \underset{\begin{subarray}{c} +3 \\ +3 \end{subarray}}{\phantom{_1\!}\mathrm{Al}\phantom{_1\!}} \underset{\begin{subarray}{c} -2 \\ -8 \end{subarray}}{(\phantom{_1\!}\mathrm{O}\phantom{_1\!}} \underset{\begin{subarray}{c} +1 \\ +4 \end{subarray}}{\phantom{_1\!}\mathrm{H}\phantom{_1\!}})_4^- \mathrm{(aq)} \longrightarrow \underset{\begin{subarray}{c} +3 \\ +3 \end{subarray}}{2~\phantom{_1\!}\mathrm{Al}\phantom{_1\!}} \underset{\begin{subarray}{c} -2 \\ -4 \end{subarray}}{\phantom{_1\!}\mathrm{O_2}\phantom{_1\!}}^-\mathrm{(aq)} ~+~ \underset{\begin{subarray}{c} +1\\ +2 \end{subarray}}{2~\phantom{_1\!}\mathrm{H_2}\phantom{_1\!}} \underset{\begin{subarray}{c} -2\\ -2 \end{subarray}}{\phantom{_1\!}\mathrm{O}\phantom{_1\!}}\mathrm{(l)} \end{align*} \]

Not a redox reaction.

Concept: oxidation state

Determine the half-reactions, the amount (in mol) of electrons transferred, and the overall balanced reaction (with phase labels) for the following redox reaction.

\[ \begin{align*} \mathrm{Fe(s)} + \mathrm{Br_2(l)} \longrightarrow \mathrm{FeBr_3(s)} \end{align*} \]

Solution

Answer:

\[ \begin{align*} \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{Fe}\phantom{_1\!}}\mathrm{(s)} ~+~ \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{Br_2}\phantom{_1\!}}\mathrm{(l)} \longrightarrow \underset{\begin{subarray}{c} +3 \\ +3 \end{subarray}}{\phantom{_1\!}\mathrm{Fe}\phantom{_1\!}} \underset{\begin{subarray}{c} -1 \\ -3 \end{subarray}}{\phantom{_1\!}\mathrm{Br_3}\phantom{_1\!}}\mathrm{(s)} \end{align*} \]

Fe undergoes oxidation and Br₂ undergoes reduction.

\[ \begin{alignat*}{3} \textrm{Ox. half-reaction} &\quad & \mathrm{Fe} &\; &\longrightarrow &\; \mathrm{Fe^{3+}} + 3~e^- \\ \textrm{Red. half-reaction} &\quad & \mathrm{Br_2} + 2~e^- &\; &\longrightarrow &\; 2~\mathrm{Br^-}\\[1.0ex] \textit{balance e}^- &\quad & 2~( \mathrm{Fe} &\; &\longrightarrow &\; \mathrm{Fe^{3+}} + 3~e^- ) \\ &\quad & 3~( \mathrm{Br_2} + 2~e^- &\; &\longrightarrow &\; 2~\mathrm{Br^-})\\[1.0ex] \textit{6 mol e}^- &\quad & 2~\mathrm{Fe} &\; &\longrightarrow &\; 2~\mathrm{Fe^{3+}} + 6~e^- \\ &\quad & 3~\mathrm{Br_2} + 6~e^- &\; &\longrightarrow &\; 6~\mathrm{Br^-} \\[1.0ex] \textit{combine} &\quad & 2~\mathrm{Fe} + 3~\mathrm{Br_2} &\; &\longrightarrow &\; 2~\mathrm{Fe^{3+}} + 6~\mathrm{Br^-} \\[1.0ex] \textrm{Balanced reaction} &\quad & 2~\mathrm{Fe(s)} + 3~\mathrm{Br_2(l)} &\; &\longrightarrow &\; 2~\mathrm{FeBr_3(s)} \end{alignat*} \]

Concept: oxidation state; oxidation-reduction reaction; balancing equations

Determine the half-reactions, the amount (in mol) of electrons transferred, and the overall balanced reaction (with phase labels) for the following redox reaction.

\[ \begin{align*} \mathrm{Mn(s)} + \mathrm{F_2(g)} \longrightarrow \mathrm{MnF_4(s)} \end{align*} \]

Solution

Answer:

\[ \begin{align*} \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{Mn}\phantom{_1\!}}\mathrm{(s)} ~+~ \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{F_2}\phantom{_1\!}}\mathrm{(g)} \longrightarrow \underset{\begin{subarray}{c} +4 \\ +4 \end{subarray}}{\phantom{_1\!}\mathrm{Mn}\phantom{_1\!}} \underset{\begin{subarray}{c} -1 \\ -1 \end{subarray}}{\phantom{_1\!}\mathrm{F_4}\phantom{_1\!}}\mathrm{(s)} \end{align*} \]

Mn undergoes oxidation and F₂ undergoes reduction.

\[ \begin{alignat*}{3} \textrm{Ox. half-reaction} &\quad & \mathrm{Mn} &\; &\longrightarrow &\; \mathrm{Mn^{4+}} + 4~e^- \\ \textrm{Red. half-reaction} &\quad & \mathrm{F_2} + 2~e^- &\; &\longrightarrow &\; 2~\mathrm{F^-}\\[1.0ex] \textit{balance e}^- &\quad & \mathrm{Mn} &\; &\longrightarrow &\; \mathrm{Mn^{4+}} + 4~e^- \\ &\quad & 2~( \mathrm{F_2} + 2~e^- &\; &\longrightarrow &\; 2~\mathrm{F^-})\\[1.0ex] \textit{4 mol e}^- &\quad & \mathrm{Mn} &\; &\longrightarrow &\; \mathrm{Mn^{4+}} + 4~e^- \\ &\quad & 2~\mathrm{F_2} + 4~e^- &\; &\longrightarrow &\; 4~\mathrm{F^-} \\[1.0ex] \textit{combine} &\quad & \mathrm{Mn} + 2~\mathrm{F_2} &\; &\longrightarrow &\; \mathrm{Mn^{4+}} + 4~\mathrm{F^-} \\[1.0ex] \textrm{Balanced reaction} &\quad & \mathrm{Mn(s)} + 2~\mathrm{F_2(g)} &\; &\longrightarrow &\; \mathrm{MnF_4(s)} \end{alignat*} \]

Concept: oxidation state; oxidation-reduction reaction; balancing equations

Determine the half-reactions, the amount (in mol) of electrons transferred, and the overall balanced reaction (with phase labels) for the following redox reaction.

\[ \begin{align*} \mathrm{Na(s)} + \mathrm{I_2(g)} \longrightarrow \mathrm{NaI(s)} \end{align*} \]

Solution

Answer:

\[ \begin{align*} \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{Na}\phantom{_1\!}}\mathrm{(s)} ~+~ \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{I_2}\phantom{_1\!}}\mathrm{(g)} \longrightarrow \underset{\begin{subarray}{c} +1 \\ +1 \end{subarray}}{\phantom{_1\!}\mathrm{Na}\phantom{_1\!}} \underset{\begin{subarray}{c} -1 \\ -1 \end{subarray}}{\phantom{_1\!}\mathrm{I}\phantom{_1\!}}\mathrm{(s)} \end{align*} \]

Na undergoes oxidation and I₂ undergoes reduction.

\[ \begin{alignat*}{3} \textrm{Ox. half-reaction} &\quad & \mathrm{Na} &\; &\longrightarrow &\; \mathrm{Na^+} + e^- \\ \textrm{Red. half-reaction} &\quad & \mathrm{I_2} + 2~e^- &\; &\longrightarrow &\; 2~\mathrm{I^-}\\[1.0ex] \textit{balance e}^- &\quad & 2~(\mathrm{Na} &\; &\longrightarrow &\; \mathrm{Na^+} + e^-) \\ &\quad & \mathrm{I_2} + 2~e^- &\; &\longrightarrow &\; 2~\mathrm{I^-}\\[1.0ex] \textit{2 mol e}^- &\quad & 2~\mathrm{Na} &\; &\longrightarrow &\; 2~\mathrm{Na^+} + 2~e^- \\ &\quad & \mathrm{I_2} + 2~e^- &\; &\longrightarrow &\; 2~\mathrm{I^-} \\[1.0ex] \textit{combine} &\quad & 2~\mathrm{Na} + \mathrm{I_2} &\; &\longrightarrow &\; 2~\mathrm{Na^+} + 2~\mathrm{I^-} \\[1.0ex] \textrm{Balanced reaction} &\quad & 2~\mathrm{Na(s)} + \mathrm{I_2(g)} &\; &\longrightarrow &\; 2~\mathrm{NaI(s)} \end{alignat*} \]

Concept: oxidation state; oxidation-reduction reaction; balancing equations

Determine the half-reactions, the amount (in mol) of electrons transferred, and the overall balanced reaction (with phase labels) for the following redox reaction.

\[ \begin{align*} \mathrm{Mg(s)} + \mathrm{Cl_2(g)} \longrightarrow \mathrm{MgCl_2(s)} \end{align*} \]

Solution

Answer:

\[ \begin{align*} \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{Mg}\phantom{_1\!}}\mathrm{(s)} ~+~ \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{Cl_2}\phantom{_1\!}}\mathrm{(g)} \longrightarrow \underset{\begin{subarray}{c} +2 \\ +2 \end{subarray}}{\phantom{_1\!}\mathrm{Mg}\phantom{_1\!}} \underset{\begin{subarray}{c} -1 \\ -1 \end{subarray}}{\phantom{_1\!}\mathrm{Cl_2}\phantom{_1\!}}\mathrm{(s)} \end{align*} \]

Mg undergoes oxidation and Cl₂ undergoes reduction.

\[ \begin{alignat*}{3} \textrm{Ox. half-reaction} &\quad & \mathrm{Mg} &\; &\longrightarrow &\; \mathrm{Mg^{2+}} + 2~e^- \\ \textrm{Red. half-reaction} &\quad & \mathrm{Cl_2} + 2~e^- &\; &\longrightarrow &\; 2~\mathrm{Cl^-}\\[1.0ex] \textit{balance e}^- &\quad & \mathrm{Mg} &\; &\longrightarrow &\; \mathrm{Mg^{2+}} + 2~e^- \\ &\quad & \mathrm{Cl_2} + 2~e^- &\; &\longrightarrow &\; 2~\mathrm{Cl^-}\\[1.0ex] \textit{2 mol e}^- &\quad & \mathrm{Mg} &\; &\longrightarrow &\; \mathrm{Mg^{2+}} + 2~e^- \\ &\quad & \mathrm{Cl_2} + 2~e^- &\; &\longrightarrow &\; 2~\mathrm{Cl^-} \\[1.0ex] \textit{combine} &\quad & \mathrm{Mg} + \mathrm{Cl_2} &\; &\longrightarrow &\; \mathrm{Mg^{2+}} + 2~\mathrm{Cl^-} \\[1.0ex] \textrm{Balanced reaction} &\quad & \mathrm{Mg(s)} + \mathrm{Cl_2(g)} &\; &\longrightarrow &\; \mathrm{MgCl_2(s)} \end{alignat*} \]

Concept: oxidation state; oxidation-reduction reaction; balancing equations

Determine the half-reactions, the amount (in mol) of electrons transferred, and the overall balanced reaction (with phase labels) for the following redox reaction.

\[ \begin{align*} \mathrm{Fe(s)} + \mathrm{O_2(g)} \longrightarrow \mathrm{Fe_2O_3(s)} \end{align*} \]

Solution

Answer:

\[ \begin{align*} \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{Fe}\phantom{_1\!}}\mathrm{(s)} ~+~ \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{O_2}\phantom{_1\!}}\mathrm{(g)} \longrightarrow \underset{\begin{subarray}{c} +3 \\ +6 \end{subarray}}{\phantom{_1\!}\mathrm{Fe_2}\phantom{_1\!}} \underset{\begin{subarray}{c} -2 \\ -6 \end{subarray}}{\phantom{_1\!}\mathrm{O_3}\phantom{_1\!}}\mathrm{(s)} \end{align*} \]

Fe undergoes oxidation and O\(_2\) undergoes reduction.

\[ \begin{alignat*}{3} \textrm{Ox. half-reaction} &\quad & 2~\mathrm{Fe} &\; &\longrightarrow &\; 2~\mathrm{Fe^{3+}} + 6~e^- \\ \textrm{Red. half-reaction} &\quad & \mathrm{O_2} + 4~e^- &\; &\longrightarrow &\; 2~\mathrm{O^{2-}} \\[1.0ex] \textit{balance e}^- &\quad & 2 ( 2~\mathrm{Fe} &\; &\longrightarrow &\; 2~\mathrm{Fe^{3+}} + 6~e^- ) \\ &\quad & 3 ( \mathrm{O_2} + 4~e^- &\; &\longrightarrow &\; 2~\mathrm{O^{2-}} ) \\[1.0ex] \textit{12 mol e}^- &\quad & 4~\mathrm{Fe} &\; &\longrightarrow &\; 4~\mathrm{Fe^{3+}} + 12~e^- \\ &\quad & 3~\mathrm{O_2} + 12~e^- &\; &\longrightarrow &\; 6~\mathrm{O^{2-}} \\[1.0ex] \textit{combine} &\quad & 4~\mathrm{Fe} + 3~\mathrm{O_2} &\; &\longrightarrow &\; 4~\mathrm{Fe^{3+}} + 6~\mathrm{O^{2-}} \\[1.0ex] \textrm{Balanced reaction} &\quad & 4~\mathrm{Fe(s)} + 3~\mathrm{O_2(g)} &\; &\longrightarrow &\; 2~\mathrm{Fe_2O_3(s)} \end{alignat*} \]

Concept: oxidation state; oxidation-reduction reaction; balancing equations

Determine the half-reactions, the amount (in mol) of electrons transferred, and the overall balanced reaction (with phase labels) for the following redox reaction.

\[ \begin{align*} \mathrm{Cr^{3+}(aq)} + \mathrm{Mn^{2+}(aq)} \longrightarrow \mathrm{Mn^{6+}(aq)} + \mathrm{Cr(s)} \end{align*} \]

Solution

Answer:

\[ \begin{align*} \underset{\begin{subarray}{c} +3\\ +3 \end{subarray}}{\phantom{_1\!}\mathrm{Cr^{3+}}\phantom{_1\!}}\mathrm{(aq)} ~+~ \underset{\begin{subarray}{c} +2\\ +2 \end{subarray}}{\phantom{_1\!}\mathrm{Mn^{2+}}\phantom{_1\!}}\mathrm{(aq)} \longrightarrow \underset{\begin{subarray}{c} +6 \\ +6 \end{subarray}}{\phantom{_1\!}\mathrm{Mn^{6+}}\phantom{_1\!}}\mathrm{(aq)} \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{Cr}\phantom{_1\!}}\mathrm{(s)} \end{align*} \]

Mn²⁺ undergoes oxidation and Cr³⁺ undergoes reduction.

\[ \begin{alignat*}{3} \textrm{Ox. half-reaction} &\quad & \mathrm{Mn^{2+}} &\; &\longrightarrow &\; \mathrm{Mn^{6+}} + 4~e^- \\ \textrm{Red. half-reaction} &\quad & \mathrm{Cr^{3+}} + 3~e^- &\; &\longrightarrow &\; \mathrm{Cr} \\[1.0ex] \textit{balance e}^- &\quad & 3(\mathrm{Mn^{2+}} &\; &\longrightarrow &\; \mathrm{Mn^{6+}} + 4~e^-) \\ &\quad & 4(\mathrm{Cr^{3+}} + 3~e^- &\; &\longrightarrow &\; \mathrm{Cr}) \\[1.0ex] \textit{12 mol e}^- &\quad & 3~\mathrm{Mn^{2+}} &\; &\longrightarrow &\; 3~\mathrm{Mn^{6+}} + 12~e^- \\ &\quad & 4~\mathrm{Cr^{3+}} + 12~e^- &\; &\longrightarrow &\; 4~\mathrm{Cr} \\[1.0ex] \textit{combine} &\quad & 4~\mathrm{Cr^{3+}} + 3~\mathrm{Mn^{2+}} &\; &\longrightarrow &\; 3~\mathrm{Mn^{6+}} + 4~\mathrm{Cr} \\[1.0ex] \textrm{Balanced reaction} &\quad & 4~\mathrm{Cr^{3+}(aq)} + 3~\mathrm{Mn^{2+}(aq)} &\; &\longrightarrow &\; 3~\mathrm{Mn^{6+}(aq)} + 4~\mathrm{Cr(s)} \end{alignat*} \]

Concept: oxidation state; oxidation-reduction reaction; balancing equations