Question

$15g$ sample of $Ba{O}_2$ is heated to ${794}^{o}C$ in a close evacuated vessel of $5$ litre capacity. How many $g$ of peroxide are converted to $BaO\left(s\right)$.
$Ba{O}_2\left(s\right)\rightleftharpoons 2BaO\left(s\right)+O_2\left(g\right),K_P=0.5atm$.

Answer 1

$K_P=\left(P_{O_2}\right)=0.5atm$

$\because PV=\frac{m}{M}RT$ or $PV=nRT$

$\therefore$ no. of moles of $O_2=n=\frac{PV}{RT}=\frac{0.5\times 5}{0.0821\times 1067}=0.028$

1 mole of $Ba{O}_2$ gives 1 mole of $O_2$

$\therefore$ 0.028 moles of $Ba{O}_2$ gives 0.028 moles of $O_2$.

$\therefore$ Weight of $Ba{O}_2$ converted to $BaO=0.028\times M_{Ba{O}_2}$

$=0.028\times 169=4.732g$