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  4. What volume of oxygen, at 18°C and 750 torr, can be obtained from 110 g of KClO3 ?

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Q. What volume of oxygen, at $ 18^{\circ}C $ and $ 750 $ torr, can be obtained from $110 \,g$ of $ KClO_3 $ ?

AMUAMU 2013States of Matter

Solution:

$\underset{2\times 122.5}{2KClO_3} \rightarrow 2KCl + \underset{3\,mol O_2}{3O_2}$
$ = 245\,g$
$245\,g\, KClO_3$ gives $ = 3\,mol\,O_2$
$110\,g \,KClO_3$will give $ = \frac{3\times 110}{245} = 1.346$ mole
$ pV = nRT$
Given, $p = 750\,torr, V = 7$
$R = 62.364\,dm^3\,Torr\,K^{-1} \,mol^{-1}$
and $T = 273 + 18 = 291 \,K$
$V = \frac{1.346 \times 62.34 \times 291}{750}$
$ = 32.569\,L \approx 32.6\,L$