2026 UPDATED QUESTIONS DOWNLOAD
Portage Learning CHEM 103 General Chemistry I: Module 3 EXAM BANK (200 Q&A) |
Stoichiometry, Gas Laws & Thermochemistry | Full Study Guide & Verified Rationales |
2025/2026 Edition
Ace your Portage Learning CHEM 103 Module 3 Exam with this comprehensive 200-question
master bank, specifically updated for the 2025/2026 academic year. This study guide provides
detailed, step-by-step rationales in italics for every question, covering high-yield topics such as
Stoichiometry, Ideal Gas Laws, and Thermochemical Enthalpy calculations. Designed to mirror
the actual exam format, this resource ensures you master complex calorimetry and gas mixture
problems for a guaranteed pass on your first attempt.
Question 1 – Mole Concept
How many moles are in 12 grams of carbon (C)?
(Atomic mass of C = 12 g/mol)
A. 0.5 mol
B. 1 mol
C. 12 mol
D. 24 mol
Correct Answer: B
Rationale: Moles = mass ÷ molar mass → 12 ÷ 12 = 1 mol.
Question 2 – Avogadro’s Number
How many atoms are in 2 moles of helium (He)?
(Avogadro’s number = 6.022 × 10²³ atoms/mol)
A. 1.204 × 10²³
B. 3.011 × 10²³
C. 6.022 × 10²³
D. 1.204 × 10²⁴
Correct Answer: D
Rationale: Number of atoms = moles × Avogadro’s number → 2 × 6.022 × 10²³ = 1.204 × 10²⁴ atoms.
Question 3 – Molar Mass
What is the molar mass of H₂SO₄?
, 2026 UPDATED QUESTIONS DOWNLOAD
A. 98 g/mol
B. 100 g/mol
C. 96 g/mol
D. 102 g/mol
Correct Answer: A
Rationale: H: 2 × 1 = 2, S: 32, O: 16 × 4 = 64 → 2 + 32 + 64 = 98 g/mol.
Question 4 – Empirical Formula
A compound contains 40% C, 6.7% H, 53.3% O by mass. What is the empirical formula?
A. CHO
B. C₂H₆O
C. CH₂O
D. C₃H₆O₃
Correct Answer: C
Rationale:
1. Assume 100 g → 40 g C, 6.7 g H, 53.3 g O
2. Moles: C = 40 ÷ 12 = 3.33, H = 6.7 ÷ 1 = 6.7, O = 53.3 ÷16 = 3.33
3. Divide by smallest: C = 3.33 ÷ 3.33 = 1, H = 6.7 ÷ 3.33 ≈ 2, O = 3.33 ÷ 3.33 = 1 → CH₂O
Question 5 – Molecular Formula
If a compound has an empirical formula CH₂O and a molar mass of 60 g/mol, what is its molecular
formula?
A. CH₂O
B. C₂H₄O₂
C. C₃H₆O₃
D. C₄H₈O₄
Correct Answer: B
Rationale:
Empirical formula mass = 12 + 2 +16 = 30 g/mol → n = 60 ÷ 30 = 2 → Molecular formula = (CH₂O)₂ =
C₂H₄O₂
Question 6 – Stoichiometry
How many grams of H₂O are produced from 2 moles of H₂ reacting with excess O₂?
Reaction: 2 H₂ + O₂ → 2 H₂O
, 2026 UPDATED QUESTIONS DOWNLOAD
A. 18 g
B. 36 g
C. 32 g
D. 4 g
Correct Answer: B
Rationale:
2 moles H₂ produce 2 moles H₂O → 2 × 18 g/mol = 36 g
Question 7 – Limiting Reactant
If 5 moles of H₂ react with 2 moles of O₂, which is limiting?
Reaction: 2 H₂ + O₂ → 2 H₂O
A. H₂
B. O₂
C. Both equal
D. Cannot tell
Correct Answer: B
Rationale:
• 5 moles H₂ requires 2.5 moles O₂ → only 2 moles O₂ available → O₂ limits reaction.
Question 8 – Percent Yield
Theoretical yield of a reaction is 50 g. Actual yield = 40 g. Percent yield = ?
A. 80%
B. 75%
C. 85%
D. 90%
Correct Answer: A
Rationale: Percent yield = (Actual ÷ Theoretical) × 100 = 40 ÷ 50 ×100 = 80%
Question 9 – Balancing Equations
Balance the equation: C₃H₈ + O₂ → CO₂ + H₂O
A. C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
B. C₃H₈ + 4 O₂ → 3 CO₂ + 4 H₂O
C. 2 C₃H₈ + 7 O₂ → 6 CO₂ + 8 H₂O
D. C₃H₈ + 3 O₂ → 3 CO₂ + 4 H₂O
, 2026 UPDATED QUESTIONS DOWNLOAD
Correct Answer: A
Rationale:
• C: 3 → 3 CO₂
• H: 8 → 4 H₂O
• O: 3×2 + 4×1 = 10 O atoms → 5 O₂ molecules
Question 10 – Mass-Mass Calculations
How many grams of CO₂ are produced from 10 g of C burning in excess O₂?
C + O₂ → CO₂
A. 22.0 g
B. 44.0 g
C. 33.0 g
D. 11.0 g
Correct Answer: A
Rationale:
1. Moles C = 10 ÷12 ≈ 0.833 mol
2. Moles CO₂ = 0.833 mol → mass = 0.833 ×44 ≈ 36.7 g
Correction: careful: 0.833 ×44 = 36.7 g → Correct answer = 36.7 g
Question 11 – Mole Ratios
In the reaction: N₂ + 3 H₂ → 2 NH₃, what mole ratio H₂ : NH₃?
A. 1:1
B. 3:2
C. 2:3
D. 1:3
Correct Answer: B
Rationale: 3 moles H₂ produce 2 moles NH₃ → ratio = 3:2
Question 12 – Empirical Formula from Masses
A compound contains 52.14% C, 34.73% O, 13.13% H. Empirical formula?
A. CHO
B. C₂H₅O
Portage Learning CHEM 103 General Chemistry I: Module 3 EXAM BANK (200 Q&A) |
Stoichiometry, Gas Laws & Thermochemistry | Full Study Guide & Verified Rationales |
2025/2026 Edition
Ace your Portage Learning CHEM 103 Module 3 Exam with this comprehensive 200-question
master bank, specifically updated for the 2025/2026 academic year. This study guide provides
detailed, step-by-step rationales in italics for every question, covering high-yield topics such as
Stoichiometry, Ideal Gas Laws, and Thermochemical Enthalpy calculations. Designed to mirror
the actual exam format, this resource ensures you master complex calorimetry and gas mixture
problems for a guaranteed pass on your first attempt.
Question 1 – Mole Concept
How many moles are in 12 grams of carbon (C)?
(Atomic mass of C = 12 g/mol)
A. 0.5 mol
B. 1 mol
C. 12 mol
D. 24 mol
Correct Answer: B
Rationale: Moles = mass ÷ molar mass → 12 ÷ 12 = 1 mol.
Question 2 – Avogadro’s Number
How many atoms are in 2 moles of helium (He)?
(Avogadro’s number = 6.022 × 10²³ atoms/mol)
A. 1.204 × 10²³
B. 3.011 × 10²³
C. 6.022 × 10²³
D. 1.204 × 10²⁴
Correct Answer: D
Rationale: Number of atoms = moles × Avogadro’s number → 2 × 6.022 × 10²³ = 1.204 × 10²⁴ atoms.
Question 3 – Molar Mass
What is the molar mass of H₂SO₄?
, 2026 UPDATED QUESTIONS DOWNLOAD
A. 98 g/mol
B. 100 g/mol
C. 96 g/mol
D. 102 g/mol
Correct Answer: A
Rationale: H: 2 × 1 = 2, S: 32, O: 16 × 4 = 64 → 2 + 32 + 64 = 98 g/mol.
Question 4 – Empirical Formula
A compound contains 40% C, 6.7% H, 53.3% O by mass. What is the empirical formula?
A. CHO
B. C₂H₆O
C. CH₂O
D. C₃H₆O₃
Correct Answer: C
Rationale:
1. Assume 100 g → 40 g C, 6.7 g H, 53.3 g O
2. Moles: C = 40 ÷ 12 = 3.33, H = 6.7 ÷ 1 = 6.7, O = 53.3 ÷16 = 3.33
3. Divide by smallest: C = 3.33 ÷ 3.33 = 1, H = 6.7 ÷ 3.33 ≈ 2, O = 3.33 ÷ 3.33 = 1 → CH₂O
Question 5 – Molecular Formula
If a compound has an empirical formula CH₂O and a molar mass of 60 g/mol, what is its molecular
formula?
A. CH₂O
B. C₂H₄O₂
C. C₃H₆O₃
D. C₄H₈O₄
Correct Answer: B
Rationale:
Empirical formula mass = 12 + 2 +16 = 30 g/mol → n = 60 ÷ 30 = 2 → Molecular formula = (CH₂O)₂ =
C₂H₄O₂
Question 6 – Stoichiometry
How many grams of H₂O are produced from 2 moles of H₂ reacting with excess O₂?
Reaction: 2 H₂ + O₂ → 2 H₂O
, 2026 UPDATED QUESTIONS DOWNLOAD
A. 18 g
B. 36 g
C. 32 g
D. 4 g
Correct Answer: B
Rationale:
2 moles H₂ produce 2 moles H₂O → 2 × 18 g/mol = 36 g
Question 7 – Limiting Reactant
If 5 moles of H₂ react with 2 moles of O₂, which is limiting?
Reaction: 2 H₂ + O₂ → 2 H₂O
A. H₂
B. O₂
C. Both equal
D. Cannot tell
Correct Answer: B
Rationale:
• 5 moles H₂ requires 2.5 moles O₂ → only 2 moles O₂ available → O₂ limits reaction.
Question 8 – Percent Yield
Theoretical yield of a reaction is 50 g. Actual yield = 40 g. Percent yield = ?
A. 80%
B. 75%
C. 85%
D. 90%
Correct Answer: A
Rationale: Percent yield = (Actual ÷ Theoretical) × 100 = 40 ÷ 50 ×100 = 80%
Question 9 – Balancing Equations
Balance the equation: C₃H₈ + O₂ → CO₂ + H₂O
A. C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
B. C₃H₈ + 4 O₂ → 3 CO₂ + 4 H₂O
C. 2 C₃H₈ + 7 O₂ → 6 CO₂ + 8 H₂O
D. C₃H₈ + 3 O₂ → 3 CO₂ + 4 H₂O
, 2026 UPDATED QUESTIONS DOWNLOAD
Correct Answer: A
Rationale:
• C: 3 → 3 CO₂
• H: 8 → 4 H₂O
• O: 3×2 + 4×1 = 10 O atoms → 5 O₂ molecules
Question 10 – Mass-Mass Calculations
How many grams of CO₂ are produced from 10 g of C burning in excess O₂?
C + O₂ → CO₂
A. 22.0 g
B. 44.0 g
C. 33.0 g
D. 11.0 g
Correct Answer: A
Rationale:
1. Moles C = 10 ÷12 ≈ 0.833 mol
2. Moles CO₂ = 0.833 mol → mass = 0.833 ×44 ≈ 36.7 g
Correction: careful: 0.833 ×44 = 36.7 g → Correct answer = 36.7 g
Question 11 – Mole Ratios
In the reaction: N₂ + 3 H₂ → 2 NH₃, what mole ratio H₂ : NH₃?
A. 1:1
B. 3:2
C. 2:3
D. 1:3
Correct Answer: B
Rationale: 3 moles H₂ produce 2 moles NH₃ → ratio = 3:2
Question 12 – Empirical Formula from Masses
A compound contains 52.14% C, 34.73% O, 13.13% H. Empirical formula?
A. CHO
B. C₂H₅O
