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PORTAGE LEARNING CHEM 103
FINAL EXAM (2025/2026) –
VERIFIED QUESTIONS FROM
EXAMS 1–6
Below are 60 verified chemistry questions covering conversions, balancing equations, gas laws,
molarity, and reactions, designed for the Portage Learning CHEM 103 Final Exam (2025/2026).
Each question includes the correct answer iand a simplified rationale for clarity.
Conversions (Questions 1–12)
1. How many grams are in 2.5 moles of NaCl?
o Answer: 146.25 g
o Rationale: Molar mass of NaCl = 58.5 g/mol (Na: 23 g/mol, Cl: 35.5 g/mol).
Mass = moles × molar mass = 2.5 mol × 58.5 g/mol = 146.25 g.
2. Convert 500 mL to liters.
o Answer: 0.5 L
o Rationale: 1 L = 1000 mL. Divide 500 mL by 1000 = 0.5 L.
3. How many moles are in 36 g of H₂O?
o Answer: 2 mol
o Rationale: Molar mass of H₂O = 18 g/mol (H: 2 × 1 g/mol, O: 16 g/mol). Moles
= mass ÷ molar mass = 36 g ÷ 18 g/mol = 2 mol.
4. Convert 25 °C to Kelvin.
o Answer: 298 K
o Rationale: K = °C + 273 = 25 + 273 = 298 K.
5. How many milliliters are in 0.75 L?
o Answer: 750 mL
o Rationale: 1 L = 1000 mL. Multiply 0.75 L × 1000 = 750 mL.
6. What is the mass of 0.2 moles of CO₂?
o Answer: 8.8 g
o Rationale: Molar mass of CO₂ = 44 g/mol (C: 12 g/mol, O: 2 × 16 g/mol). Mass
= 0.2 mol × 44 g/mol = 8.8 g.
7. Convert 1.5 kg to grams.
o Answer: 1500 g
o Rationale: 1 kg = 1000 g. Multiply 1.5 kg × 1000 = 1500 g.
8. How many moles are in 112 g of Fe?
o Answer: 2 mol
o Rationale: Molar mass of Fe = 56 g/mol. Moles = 112 g ÷ 56 g/mol = 2 mol.
9. Convert 760 mmHg to atm.
, 2
o Answer: 1 atm
o Rationale: 1 atm = 760 mmHg. Divide 760 mmHg by 760 = 1 atm.
10. How many liters are in 2500 cm³?
o Answer: 2.5 L
o Rationale: 1 L = 1000 cm³. Divide 2500 cm³ by 1000 = 2.5 L.
11. What is the mass of 3 moles of NH₃?
o Answer: 51 g
o Rationale: Mar mass of NH₃ = 17 g/mol (N: 14 g/mol, H: 3 × 1 g/mol). Mass = 3
mol × 17 g/mol = 51 g.
12. Convert 0.0821 L·atm/(mol·K) to mL·mmHg/(mol·K).
o Answer: 62.4 mL·mmHg/(mol·K)
o Rationale: 1 L = 1000 mL, 1 atm = 760 mmHg. Multiply 0.0821 × 1000 × 760 =
62.4 mL·mmHg/(mol·K).
Balancing Equations (Questions 13–24)
13. Balance: __ C₂H₅OH + __ O₂ → __ CO₂ + __ H₂O
o Answer: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
o Rationale: Balance C (2 on left, 2CO₂ on right), H (6 on left, 3H₂O on right), O
(7 on right, 1 + 3O₂ = 7 on left).
14. Balance: __ Na + __ H₂O → __ NaOH + __ H₂
o Answer: 2Na + 2H₂O → 2NaOH + H₂
o Rationale: Balance Na (2 on left, 2NaOH on right), H (4 on left, 2NaOH + H₂ on
right), O (2 on left, 2 on right).
15. Balance: __ CH₄ + __ O₂ → __ CO₂ + __ H₂O
o Answer: CH₄ + 2O₂ → CO₂ + 2H₂O
o Rationale: Balance C (1 on each side), H (4 on left, 2H₂O on right), O (4 on right,
2O₂ on left).
16. Balance: __ Fe + __ O₂ → __ Fe₂O₃
o Answer: 4Fe + 3O₂ → 2Fe₂O₃
o Rationale: Balance Fe (4 on left, 2Fe₂O₃ on right), O (6 on right, 3O₂ on left).
17. Balance: __ Al + __ HCl → __ AlCl₃ + __ H₂
o Answer: 2Al + 6HCl → 2AlCl₃ + 3H₂
o Rationale: Balance Al (2 on left, 2AlCl₃ on right), Cl (6 on left, 6 on right), H (6
on left, 3H₂ on right).
18. Balance: __ N₂ + __ H₂ → __ NH₃
o Answer: N₂ + 3H₂ → 2NH₃
o Rationale: Balance N (2 on left, 2NH₃ on right), H (6 on left, 6 on right).
19. Balance: __ KClO₃ → __ KCl + __ O₂
o Answer: 2KClO₃ → 2KCl + 3O₂
o Rationale: Balance K (2 on left, 2KCl on right), Cl (2 on left, 2 on right), O (6 on
left, 3O₂ on right).
20. Balance: __ C₃H₈ + __ O₂ → __ CO₂ + __ H₂O
o Answer: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
PORTAGE LEARNING CHEM 103
FINAL EXAM (2025/2026) –
VERIFIED QUESTIONS FROM
EXAMS 1–6
Below are 60 verified chemistry questions covering conversions, balancing equations, gas laws,
molarity, and reactions, designed for the Portage Learning CHEM 103 Final Exam (2025/2026).
Each question includes the correct answer iand a simplified rationale for clarity.
Conversions (Questions 1–12)
1. How many grams are in 2.5 moles of NaCl?
o Answer: 146.25 g
o Rationale: Molar mass of NaCl = 58.5 g/mol (Na: 23 g/mol, Cl: 35.5 g/mol).
Mass = moles × molar mass = 2.5 mol × 58.5 g/mol = 146.25 g.
2. Convert 500 mL to liters.
o Answer: 0.5 L
o Rationale: 1 L = 1000 mL. Divide 500 mL by 1000 = 0.5 L.
3. How many moles are in 36 g of H₂O?
o Answer: 2 mol
o Rationale: Molar mass of H₂O = 18 g/mol (H: 2 × 1 g/mol, O: 16 g/mol). Moles
= mass ÷ molar mass = 36 g ÷ 18 g/mol = 2 mol.
4. Convert 25 °C to Kelvin.
o Answer: 298 K
o Rationale: K = °C + 273 = 25 + 273 = 298 K.
5. How many milliliters are in 0.75 L?
o Answer: 750 mL
o Rationale: 1 L = 1000 mL. Multiply 0.75 L × 1000 = 750 mL.
6. What is the mass of 0.2 moles of CO₂?
o Answer: 8.8 g
o Rationale: Molar mass of CO₂ = 44 g/mol (C: 12 g/mol, O: 2 × 16 g/mol). Mass
= 0.2 mol × 44 g/mol = 8.8 g.
7. Convert 1.5 kg to grams.
o Answer: 1500 g
o Rationale: 1 kg = 1000 g. Multiply 1.5 kg × 1000 = 1500 g.
8. How many moles are in 112 g of Fe?
o Answer: 2 mol
o Rationale: Molar mass of Fe = 56 g/mol. Moles = 112 g ÷ 56 g/mol = 2 mol.
9. Convert 760 mmHg to atm.
, 2
o Answer: 1 atm
o Rationale: 1 atm = 760 mmHg. Divide 760 mmHg by 760 = 1 atm.
10. How many liters are in 2500 cm³?
o Answer: 2.5 L
o Rationale: 1 L = 1000 cm³. Divide 2500 cm³ by 1000 = 2.5 L.
11. What is the mass of 3 moles of NH₃?
o Answer: 51 g
o Rationale: Mar mass of NH₃ = 17 g/mol (N: 14 g/mol, H: 3 × 1 g/mol). Mass = 3
mol × 17 g/mol = 51 g.
12. Convert 0.0821 L·atm/(mol·K) to mL·mmHg/(mol·K).
o Answer: 62.4 mL·mmHg/(mol·K)
o Rationale: 1 L = 1000 mL, 1 atm = 760 mmHg. Multiply 0.0821 × 1000 × 760 =
62.4 mL·mmHg/(mol·K).
Balancing Equations (Questions 13–24)
13. Balance: __ C₂H₅OH + __ O₂ → __ CO₂ + __ H₂O
o Answer: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
o Rationale: Balance C (2 on left, 2CO₂ on right), H (6 on left, 3H₂O on right), O
(7 on right, 1 + 3O₂ = 7 on left).
14. Balance: __ Na + __ H₂O → __ NaOH + __ H₂
o Answer: 2Na + 2H₂O → 2NaOH + H₂
o Rationale: Balance Na (2 on left, 2NaOH on right), H (4 on left, 2NaOH + H₂ on
right), O (2 on left, 2 on right).
15. Balance: __ CH₄ + __ O₂ → __ CO₂ + __ H₂O
o Answer: CH₄ + 2O₂ → CO₂ + 2H₂O
o Rationale: Balance C (1 on each side), H (4 on left, 2H₂O on right), O (4 on right,
2O₂ on left).
16. Balance: __ Fe + __ O₂ → __ Fe₂O₃
o Answer: 4Fe + 3O₂ → 2Fe₂O₃
o Rationale: Balance Fe (4 on left, 2Fe₂O₃ on right), O (6 on right, 3O₂ on left).
17. Balance: __ Al + __ HCl → __ AlCl₃ + __ H₂
o Answer: 2Al + 6HCl → 2AlCl₃ + 3H₂
o Rationale: Balance Al (2 on left, 2AlCl₃ on right), Cl (6 on left, 6 on right), H (6
on left, 3H₂ on right).
18. Balance: __ N₂ + __ H₂ → __ NH₃
o Answer: N₂ + 3H₂ → 2NH₃
o Rationale: Balance N (2 on left, 2NH₃ on right), H (6 on left, 6 on right).
19. Balance: __ KClO₃ → __ KCl + __ O₂
o Answer: 2KClO₃ → 2KCl + 3O₂
o Rationale: Balance K (2 on left, 2KCl on right), Cl (2 on left, 2 on right), O (6 on
left, 3O₂ on right).
20. Balance: __ C₃H₈ + __ O₂ → __ CO₂ + __ H₂O
o Answer: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
