Stoichiometry -

1. When 4 moles of solid potassium chlorate (KClO3) are heated until it fully decomposes, how many moles of oxygen gas (O₂) are produced?
2KClO₃ → 2KCl + 3O₂

A. 5 mole
B. 2 moles
C. 12 moles
D. 6 moles

Answer

Explanation
Option A: According to the balanced equation, 4 moles of KClO₃ do not produce 5 moles of O₂. The correct amount is 6 moles.
Option B: This is much less than the amount predicted by the balanced chemical equation.
Option C: This value is twice the correct amount, 4 moles of KClO₃ cannot produce 12 moles of O₂.
Option D: 2 moles of KClO₃ produce 3 moles of O₂. Therefore, 4 moles of KClO₃ produce: 3×4/2 + 6 moles O₂. (Correct)

2. Which of the following samples contains exactly one mole of molecules?

A. 4 gram of hydrogen gas (H₂)
B. 17 gram of ammonia (NH₃)
C. 14 gram of nitrogen gas (N₂)
D. 88 gram of carbon dioxide (CO₂)

Answer

Explanation
Option A: The molar mass of H₂ is 2 g/mol, so 4 grams equals 2 moles, not 1 mole.
Option B: The molar mass of NH₃is 17 g/mol, so 17 grams equals exactly 1 mole of NH₃ moleculest. (Correct)
Option C: The molar mass of N₂ is 28 g/mol, so 14 grams equals only 0.5 mole nitrogen gas, not 1 mole.
Option D: The molar mass of CO₂ is 44 g/mol, so 88 grams equals only 2 mole carbon dioxide gas, not 1 mole.

3. A rocket engine burns methane in oxygen as shown by the following chemical equation.
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
If 32 g of methane is completely burned, what mass of oxygen gas is required?

A. 32 g
B. 64 g
C. 128 g
D. 256 g

Answer

Explanation
Option A: The reaction requires two moles of oxygen for every mole of methane, so equal masses are not sufficient.
Option B: Although 64 g of oxygen equals 2 moles, 32 g of methane equals 2 moles and therefore requires 4 moles of oxygen.
Option C: Since methane has a molar mass of 16 g/mol, 32 g is 2 moles. The 1:2 ratio means 2 moles of methane require 4 moles of oxygen. Thus, the required mass is 4×32=128 g. (Correct)
Option D: This is twice the required amount of oxygen.

4. What mass of nitrogen is present in 0.30 mole of ammonium nitrate (NH₄NO₃)?

A. 4.2 g
B. 8.4 g
C. 14.0 g
D. 16.8 g

Answer

Explanation
Option A: This calculation accounts for only one nitrogen atom per formula unit, but ammonium nitrate contains two.
Option B: Each NH₄NO₃ formula unit contains 2 nitrogen atoms. Therefore, 0.30 mole of NH₄NO₃ contains 0.60 mole of nitrogen atoms. The mass is $0.60 \times 14 = 8.4$ 0.60×14=8.4 g. (Correct)
Option C: This is the mass of one mole of nitrogen atoms, not the nitrogen in 0.30 mole of ammonium nitrate.
Option D: This is twice the required nitrogen mass.

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