Question

Oxygen gas generated by the decomposition of potassium chlorate is collected. The volume of oxygen collected at ${24}^{0}C$ and the atmospheric pressure of 760 mmHg is 128 mL. Calculate the mass (in grams) of oxygen gas obtained. The pressure of the water vapor at ${24}^{0}C$ is 22.4 mmHg.

• A. 1.36 g
• B. 1.52 g
• C. 0.163 g
• D. 1.63 g

Answer 1

The correct option is C 0.163 g

From the total pressure and the vapour pressure of water we can calculate the partial pressure of $O_2$.

$P{O}_2=P_t-P_{H_{2}O}$

= 760 − 22.4

= 737.6 mmHg

From the ideal gas equation we write.

$W=\frac{PVM}{RT}=\frac{\left(0.974atm\right)\left(0.128L\right)\left(32.0g/mol/\right)}{\left(0.0821Latm/Kmol\right)(273+24)K}=0.163g$.