What volume of oxygen gas, at 533°C and 750 torr, could be produced by the decomposition of 38.1 g of potassium chlorate? The other product is potassium chloride.

Question

Quizplus · Accepted Answer

1. Write the balanced chemical equation for the decomposition reaction:
2KClO3(s) → 2KCl(s) + 3O2(g)

2. Calculate the moles of KClO3:
38.1 g KClO3 × (1 mol KClO3/122.54 g KClO3) = 0.311 mol KClO3

3. Use stoichiometry to find the moles of O2 produced:
0.311 mol KClO3 × (3 mol O2/2 mol KClO3) = 0.4675 mol O2

4. Use the ideal gas law to find the volume of O2 produced:
V = (nRT)/P
V = (0.4675 mol)(0.0821 L•atm/mol•K)(533+273 K)/(750 torr)
V = 31.3 L

Therefore, the answer is B: 31.3 L.

What Volume of Oxygen Gas, at 533°C and 750 Torr