Question

When 400 gram KCLO3 is heated,10 litre of O2 at 1 ATM and at 327 degree celsius is obtained what percentage of KCL O3 is decomposed?

Answer 1

Answer: 4.08%

Explanation:

According to the ideal gas equation:'

$PV=nRT$

Pressure of the gas = 1 atm

Volume of the gas = 10 L

Temperature of the gas = 327°C = (327+273)K = 600K  (0°C = 273 K)

Value of gas constant in in kilopascals = 0.0821 Latm/K mol

$n=\frac{PV}{RT}=\frac{1atm\times 10L}{0.0821\times 600}=0.20mol$

$2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)$

3 moles of $O_2$  is produced from 2 moles of $KClO_3$

Thus 0.20 moles of $O_2$ is produced from =$\frac{2}{3}\times 0.20=0.13moles$ of $KClO_3$

Mass of $KClO_3=moles\times {\text {molar mass}}=0.13\times 122.5=16.3g$

Thus percent of $KClO_3$ decomposed is =$\frac{16.3}{400}\times 100=4.08%$

4.08 % of  $KClO_3$ decomposed.