Question

The volume of oxygen gas liberated at $27{}^{o}C$ and 760 mm Hg pressure when 24.5 g of $\text{KCl}{\text{O}}_{\text{3}}$ is heated, is $\text{(M}\text{.wt}\text{.of KC1}{\text{O}}_{\text{3}}\text{=122}\text{.5)}$

• A. 4.48 L
• B. 9.6 L
• C. 7.4 L
• D. 11.2 L

Answer 1

Answer: (C)

$\begin{array}{rl}& \underset{\begin{array}{c}2\times 122.5\\ =245.0\end{array}}{KCl{O}_3}\to \underset{\begin{array}{c}3\times 22.4\\ =67.2L\end{array}}{2KCl+3O_2}\\ & atSTP\end{array}$ $\because$ When $245gKCl{O}_3$ is heated, liberated oxygen = 67.2 L $\therefore$ When $24.5gKCl{O}_3$ is heated, liberated oxygen $=\frac{67.2\times 24.5}{245}=6.72L$ This volume of oxygen was liberated at STP. Therefore, at ${27}^{o}C$ and 760 mm Hg pressure, the volume of liberated oxygen will be (STP) (Given conditions) $\frac{p_1{V}_1}{T_1}=\frac{p_2{V}_2}{T_2}$ $\frac{760\times 6.72}{273}=\frac{760\times V_2}{300}$ $\therefore$ $V_2=\frac{760\times 6.72\times 300}{760\times 273}$ $\text{=}\text{7}\text{.38}\text{L}$ $\approx \text{7}\text{.4}\text{L}$