Question

A 3.25 g mixture of KCl and KClO3 is heated until all of the KClO3 decomposes according to the following reaction:
2KClO3(s) → 2KCl(s) + 3O2(g)
The O2 is collected and is found to have a volume of 98.0 mL at 21.0 °C and 1.00 atm. What is the percentage of KClO3 (by mass) in the mixture?

Answer 1

T = 21.0 °C = 294.15 K

P = 1 atm

V = 98.0 mL = 0.098 L

Ideal gas law:

PV=nRT

Molar gas constant

R=0.082058 L×atm/(K×mol)

$1\;atm \times 0.098\;L = n\;mol \times 0.082058 \;(L×atm/(K×mol)) \times 294.15\;K$

$0.098 = n \times 24.137$

n(O₂) = 0.004 mol

Proportion from the reaction:

2 – 3

x – 0.004 mol

x = 0.0026 mole (KClO₃)

$m = n\times M$

M(KClO₃) = 122.55 g/mol

$m(KClO_3) = 0.0026\times 122.55 = 0.318\; g$ (mass of reacted KClO₃)

Proportion:

3.25 g – 100 %

0.318 g – y

y = 9.78 %

Answer: 9.78 %.