- 2
Solid potassium chlorate decomposes to produce solid
potassium chloride and oxygen gas according to the balanced
equation: 2 KCIO
2 KCl + 3 O
What volume of oxygen gas, measured at 40.0°C and 85.4 kPa,
will be produced when 13.5 g of potassium chlorate is
decomposed?

5 × 10^-4 L

Explanation:

The equation of the reaction is;

2KClO3 = 2KCl + 3O2

Number of moles of KClO3 = 13.5g/122.5 g / mol = 0.11 moles

From the stoichiometry of the reaction;

2 moles of KClO3 yields 3 moles of O2

0.11 moles of KClO3 yields 0.11 × 3/2 = 0.165 moles of oxygen gas

From the ideal gas equation;

PV= nRT

P= 85.4 × 10^4 KPa

V=?

n= 0.165

R= 8.314 J K-1 mol-1

T= 40+273 = 313K

V= 0.165 ×8.134 × 313/85.4 × 10^4

V=429.4/85.4 × 10^4

V= 5 × 10^-4 L

fe 55.85 g/mol and h2o 18.02 g/mol.