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.
A)35.9 L
B)31.3 L
C)30.4 L
D)20.7 L
E)10.4 L

Question

Quizplus · Accepted Answer

First, write the balanced chemical equation:

2KClO3(s) → 2KCl(s) + 3O2(g)

Next, use stoichiometry to determine the amount of oxygen produced from 38.1 g of KClO3:

38.1 g KClO3 x (1 mol KClO3/122.55 g) x (3 mol O2/2 mol KClO3) = 0.929 mol O2

Then, use the ideal gas law to calculate the volume of the oxygen gas at the given conditions:

PV = nRT

V = (nRT)/P

V = [(0.929 mol)(0.08206 L•atm/mol•K)(533+273 K)] / (750 Torr)

V = 31.3 L

Therefore, the answer is B.

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