What is the mass of $\mathrm{KClO}_{3(\mathrm{~s})}$ required to liberate 22. $4 \mathrm{dm}^3$ oxygen at…
What is the mass of $\mathrm{KClO}_{3(\mathrm{~s})}$ required to liberate 22. $4 \mathrm{dm}^3$ oxygen at STP during thermal decomposition?
(Molar Mass of $\mathrm{KClO}_{3(\mathrm{~s})}=122.5 \mathrm{~g} / \mathrm{mol}$ )
$122.5 \mathrm{~g}$
$81.67 \mathrm{~g}$
$10.25 \mathrm{~g}$
$8.16 \mathrm{~g}$
Solution
$\begin{aligned}
& 2 \mathrm{KClO}_3 \\
& {[2 \mathrm{moles}]}
\end{aligned} \longrightarrow 2 \mathrm{KCl}+\underset{[3 \mathrm{moles}]}{3 \mathrm{O}_2 \uparrow}$
[3 moles]
$2 \text { moles of } \mathrm{KClO}_3=2 \times 122.5=245 \mathrm{~g}$
3 moles of $\mathrm{O}_2$ at STP occupy $=\left(3 \times 22.4 \mathrm{dm}^3\right)$.
Thus, $245 \mathrm{~g}$ of potassium chlorate will liberate $67.2 \mathrm{dm}^3$ of oxygen gas.
Let ' $x$ ' gram of $\mathrm{KClO}_3$ liberate $22.4 \mathrm{dm}^3$ of oxygen gas at S.T.P.
$\therefore \quad x=\frac{245 \times 22.4}{3 \times 22.4}=81.67 \mathrm{~g}$