Question

What volume of oxygen, at $18^{\circ }C$ and $750$ torr, can be obtained from $110g$ of $KCl{O}_3$ ?

• A. 32.6 L
• B. 42.7 L
• C. 3.26 L
• D. 4.27 L

Answer 1

Answer: (A)

$\underset{2\times 122.5}{2KCl{O}_3}\to 2KCl+\underset{3mol{O}_2}{3O_2}$
$=245g$
$245gKCl{O}_3$ gives $=3mol{O}_2$
$110gKCl{O}_3$will give $=\frac{3\times 110}{245}=1.346$ mole
$pV=nRT$
Given, $p=750torr,V=7$
$R=62.364d{m}^{3}Torr{K}^{-1}mo{l}^{-1}$
and $T=273+18=291K$
$V=\frac{1.346\times 62.34\times 291}{750}$
$=32.569L\approx 32.6L$