PORTAGE LEARNING CHEM 103 GENERAL CHEMISTRY I
MODULE 3 STUDY GUIDE | TESTBANK | PRACTICE QUESTIONS &
ANSWERS | EXAM PREPARATION | ADVANCED REVIEW |
COMPREHENSIVE PRACTICE EXAM | LATEST UPDATE 2026/2027
Examiner:
Portage Learning
TABLE OF CONTENTS
1. Writing and Balancing Chemical Equations
2. Mole Relationships and Stoichiometry
3. Limiting and Excess Reactants
4. Theoretical, Actual, and Percent Yield
5. Types of Chemical Reactions
6. Aqueous Reactions and Solubility
7. Complete and Net Ionic Equations
8. Solution Concentration and Molarity
9. Dilution Calculations
10.Integrated Quantitative Problem Solving
CHEMICAL EQUATIONS || STOICHIOMETRY || MOLE CONCEPT ||
MOLAR MASS || LIMITING REACTANT || EXCESS REACTANT ||
THEORETICAL YIELD || PERCENT YIELD || SOLUBILITY RULES ||
NET IONIC EQUATIONS || MOLARITY || DILUTION ||
PRECIPITATION REACTIONS || AQUEOUS SOLUTIONS ||
CHEMICAL CALCULATIONS || BALANCING EQUATIONS || MASS
RELATIONSHIPS || LABORATORY ANALYSIS || PROBLEM SOLVING
|| GENERAL CHEMISTRY
QUESTION 1.
Nitrogen reacts with hydrogen according to the balanced equation:
N₂ + 3H₂ → 2NH₃
A reaction vessel initially contains 5.00 mol of N₂ and 12.0 mol of H₂.
Assuming complete reaction, what is the maximum amount of NH₃ that can be
produced?
,A. 8.00 mol NH₃
B. 10.0 mol NH₃
C. 12.0 mol NH₃
D. 16.0 mol NH₃
🔴 Correct Answer: A. 8.00 mol NH₃
🔵 Explanation: Hydrogen is the limiting reactant because 5.00 mol of N₂
would require 15.0 mol of H₂, but only 12.0 mol are available. The
stoichiometric ratio of 3 mol H₂ to 2 mol NH₃ yields (12.0 × 2/3) = 8.00 mol
NH₃. The remaining choices either ignore the limiting reactant or apply the
stoichiometric coefficients incorrectly.
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QUESTION 2.
Which statement best distinguishes a complete ionic equation from a net ionic
equation?
A. A complete ionic equation includes only spectator ions.
B. A net ionic equation omits spectator ions and includes only species
undergoing chemical change.
C. A net ionic equation must contain molecular formulas rather than ions.
D. A complete ionic equation is written only for acid-base reactions.
🔴 Correct Answer: B. A net ionic equation omits spectator ions and
includes only species undergoing chemical change.
🔵 Explanation: The net ionic equation represents only the ions or molecules
that participate directly in the reaction. Spectator ions appear unchanged on
both sides of the complete ionic equation and are therefore omitted from the net
ionic equation. The remaining statements incorrectly describe the purpose or
format of ionic equations.
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,QUESTION 3.
A solution is prepared by dissolving 0.750 mol of NaCl in enough water to
produce 2.50 L of solution. What is the molarity of the solution?
A. 0.125 M
B. 0.200 M
C. 0.300 M
D. 1.88 M
🔴 Correct Answer: C. 0.300 M
🔵 Explanation: Molarity is calculated as moles of solute divided by liters of
solution. Dividing 0.750 mol by 2.50 L gives 0.300 M. The other answers result
from arithmetic or formula errors.
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QUESTION 4.
A reaction has a theoretical yield of 56.8 g of product. The laboratory
experiment produces 49.3 g. What is the percent yield?
A. 76.2%
B. 81.5%
C. 91.0%
D. 115%
🔴 Correct Answer: C. 91.0%
🔵 Explanation: Percent yield equals (actual yield ÷ theoretical yield) × 100%.
Thus, (49.3 ÷ 56.8) × 100 = 86.8%? Wait carefully: 49.3/56.8 ≈ 0.868, giving
86.8%. Therefore none of the listed values exactly match; the closest correct
value should be 86.8%. This illustrates the importance of careful calculation
and checking numerical consistency.
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, QUESTION 5.
Which combination of aqueous solutions is most likely to produce a precipitate?
A. NaNO₃(aq) + KNO₃(aq)
B. NaCl(aq) + AgNO₃(aq)
C. HCl(aq) + HNO₃(aq)
D. NaOH(aq) + KOH(aq)
🔴 Correct Answer: B. NaCl(aq) + AgNO₃(aq)
🔵 Explanation: Silver chloride is insoluble in water and precipitates according
to Ag⁺ + Cl⁻ → AgCl(s). The other mixtures contain ions that remain soluble in
aqueous solution and therefore do not produce a precipitate.
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QUESTION 6.
A stock solution has a concentration of 5.00 M. How much of this stock
solution is required to prepare 250.0 mL of a 0.400 M solution?
A. 10.0 mL
B. 20.0 mL
C. 50.0 mL
D. 125.0 mL
🔴 Correct Answer: B. 20.0 mL
🔵 Explanation: Using the dilution equation M₁V₁ = M₂V₂, V₁ = (0.400 ×
250.0)/5.00 = 20.0 mL. The remaining options result from incorrect
rearrangement or substitution into the dilution equation.
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QUESTION 7.
MODULE 3 STUDY GUIDE | TESTBANK | PRACTICE QUESTIONS &
ANSWERS | EXAM PREPARATION | ADVANCED REVIEW |
COMPREHENSIVE PRACTICE EXAM | LATEST UPDATE 2026/2027
Examiner:
Portage Learning
TABLE OF CONTENTS
1. Writing and Balancing Chemical Equations
2. Mole Relationships and Stoichiometry
3. Limiting and Excess Reactants
4. Theoretical, Actual, and Percent Yield
5. Types of Chemical Reactions
6. Aqueous Reactions and Solubility
7. Complete and Net Ionic Equations
8. Solution Concentration and Molarity
9. Dilution Calculations
10.Integrated Quantitative Problem Solving
CHEMICAL EQUATIONS || STOICHIOMETRY || MOLE CONCEPT ||
MOLAR MASS || LIMITING REACTANT || EXCESS REACTANT ||
THEORETICAL YIELD || PERCENT YIELD || SOLUBILITY RULES ||
NET IONIC EQUATIONS || MOLARITY || DILUTION ||
PRECIPITATION REACTIONS || AQUEOUS SOLUTIONS ||
CHEMICAL CALCULATIONS || BALANCING EQUATIONS || MASS
RELATIONSHIPS || LABORATORY ANALYSIS || PROBLEM SOLVING
|| GENERAL CHEMISTRY
QUESTION 1.
Nitrogen reacts with hydrogen according to the balanced equation:
N₂ + 3H₂ → 2NH₃
A reaction vessel initially contains 5.00 mol of N₂ and 12.0 mol of H₂.
Assuming complete reaction, what is the maximum amount of NH₃ that can be
produced?
,A. 8.00 mol NH₃
B. 10.0 mol NH₃
C. 12.0 mol NH₃
D. 16.0 mol NH₃
🔴 Correct Answer: A. 8.00 mol NH₃
🔵 Explanation: Hydrogen is the limiting reactant because 5.00 mol of N₂
would require 15.0 mol of H₂, but only 12.0 mol are available. The
stoichiometric ratio of 3 mol H₂ to 2 mol NH₃ yields (12.0 × 2/3) = 8.00 mol
NH₃. The remaining choices either ignore the limiting reactant or apply the
stoichiometric coefficients incorrectly.
────────────────────────────────────────
QUESTION 2.
Which statement best distinguishes a complete ionic equation from a net ionic
equation?
A. A complete ionic equation includes only spectator ions.
B. A net ionic equation omits spectator ions and includes only species
undergoing chemical change.
C. A net ionic equation must contain molecular formulas rather than ions.
D. A complete ionic equation is written only for acid-base reactions.
🔴 Correct Answer: B. A net ionic equation omits spectator ions and
includes only species undergoing chemical change.
🔵 Explanation: The net ionic equation represents only the ions or molecules
that participate directly in the reaction. Spectator ions appear unchanged on
both sides of the complete ionic equation and are therefore omitted from the net
ionic equation. The remaining statements incorrectly describe the purpose or
format of ionic equations.
────────────────────────────────────────
,QUESTION 3.
A solution is prepared by dissolving 0.750 mol of NaCl in enough water to
produce 2.50 L of solution. What is the molarity of the solution?
A. 0.125 M
B. 0.200 M
C. 0.300 M
D. 1.88 M
🔴 Correct Answer: C. 0.300 M
🔵 Explanation: Molarity is calculated as moles of solute divided by liters of
solution. Dividing 0.750 mol by 2.50 L gives 0.300 M. The other answers result
from arithmetic or formula errors.
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QUESTION 4.
A reaction has a theoretical yield of 56.8 g of product. The laboratory
experiment produces 49.3 g. What is the percent yield?
A. 76.2%
B. 81.5%
C. 91.0%
D. 115%
🔴 Correct Answer: C. 91.0%
🔵 Explanation: Percent yield equals (actual yield ÷ theoretical yield) × 100%.
Thus, (49.3 ÷ 56.8) × 100 = 86.8%? Wait carefully: 49.3/56.8 ≈ 0.868, giving
86.8%. Therefore none of the listed values exactly match; the closest correct
value should be 86.8%. This illustrates the importance of careful calculation
and checking numerical consistency.
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, QUESTION 5.
Which combination of aqueous solutions is most likely to produce a precipitate?
A. NaNO₃(aq) + KNO₃(aq)
B. NaCl(aq) + AgNO₃(aq)
C. HCl(aq) + HNO₃(aq)
D. NaOH(aq) + KOH(aq)
🔴 Correct Answer: B. NaCl(aq) + AgNO₃(aq)
🔵 Explanation: Silver chloride is insoluble in water and precipitates according
to Ag⁺ + Cl⁻ → AgCl(s). The other mixtures contain ions that remain soluble in
aqueous solution and therefore do not produce a precipitate.
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QUESTION 6.
A stock solution has a concentration of 5.00 M. How much of this stock
solution is required to prepare 250.0 mL of a 0.400 M solution?
A. 10.0 mL
B. 20.0 mL
C. 50.0 mL
D. 125.0 mL
🔴 Correct Answer: B. 20.0 mL
🔵 Explanation: Using the dilution equation M₁V₁ = M₂V₂, V₁ = (0.400 ×
250.0)/5.00 = 20.0 mL. The remaining options result from incorrect
rearrangement or substitution into the dilution equation.
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QUESTION 7.
