What volume of oxygen gas is released at STP if 10.0 g of potassium chlorate is decomposed? (The molar mass of KClO3 is 122.55 g/mol.)

Given:

10g of potassium chlorate decomposed

STP

molar mass of KClO3 is 122.55 g/mol

Required:

Volume of oxygen gas

Solution:

The decomposition reaction is

2KClO3 → 2KCl + 3O2

Moles of O2 = 10g KClO3 (1 mol KClO3/122.55 g/mol KClO3) (3 moles O2/2 moles KClO3) = 15 moles O2

Using ideal gas law: PV = nRT

PV = nRT

V = nRT/P

V = (15 moles O2) (0.08206 L-atm/mol-K) (273K) / 1 atm

V = 336.04 L O2