Question

2g of a mixture of KCLO3 and KCL yielded 336cm^3 of O2 measured dry at 747mmHg on heating. Calculate the volume of dry O2 measured at stp and the percentage composition of KCLO3.

Answer 1

2KClO₃ → 2KCl + 3O₂

We will use the temperature value of 127 ºC.

T = 127 + 273 = 400 K

P = 747 mmHg = 0.982 atm

V = 336 mL = 0.336 L

Ideal Gas Law

PV = nRT

R = 0.08206 L×atm/mol×K

$n = \frac{PV}{RT} \\ n(O_2) = \frac{0.982 \times 0.336}{0.08206 \times 400} \\ = 0.01 \;mol$

At STP one mole of any gas occupies volume of 22.4 L.

At STP $V(O_2) = 0.01 \times 22.4 = 0.224 \;L = 224 \;mL$

According to the reaction:

n(KClO₃) $= \frac{2}{3}n(O_2) = \frac{2}{3}0.01= 0.0066 \;mol$

M(KClO₃) = 122.55 g/mol

$m(KClO_3) = n \times M = 0.0066 \times 122.55 = 0.817 \;g$

Proportion:

2 g – 100 %

0.817 g – x

$x = \frac{0.817 \times 100}{2} = 40.8 \; \%$