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In the reaction
$2 \mathrm{KClO}_{3(\mathrm{~s})} \longrightarrow 2 \mathrm{KCl}_{(\mathrm{s})}+3 \mathrm{O}_{2(\mathrm{~g})} \Delta \mathrm{H}^{\circ}=-78 \mathrm{~kJ} .$ If $33 \cdot 6 \mathrm{~L}$ of oxygen gas is liberated at S.T.P.
What is the mass of $\mathrm{KCl}_{(s)}$ produced $?$ (at. mass $\mathrm{K}=39, \mathrm{Cl}=35 \cdot 5 \mathrm{~g} \mathrm{~mol}^{-1}$ )
Options:
$2 \mathrm{KClO}_{3(\mathrm{~s})} \longrightarrow 2 \mathrm{KCl}_{(\mathrm{s})}+3 \mathrm{O}_{2(\mathrm{~g})} \Delta \mathrm{H}^{\circ}=-78 \mathrm{~kJ} .$ If $33 \cdot 6 \mathrm{~L}$ of oxygen gas is liberated at S.T.P.
What is the mass of $\mathrm{KCl}_{(s)}$ produced $?$ (at. mass $\mathrm{K}=39, \mathrm{Cl}=35 \cdot 5 \mathrm{~g} \mathrm{~mol}^{-1}$ )
Solution:
2213 Upvotes
Verified Answer
The correct answer is:
$48 \cdot 0 \mathrm{~g}$
$\begin{aligned} 2 \mathrm{KClO}_{3(s)} \longrightarrow & 2 \mathrm{KCl}_{(s)}+\quad 3 \mathrm{O}_{2(s)} \\ & 2 \times 74.5 \mathrm{~g} \quad 3 \times 22.4 \mathrm{~L} \\ &=149 \mathrm{~g} \quad=67.2 \mathrm{~L} \end{aligned}$
Now, $67.2 \mathrm{~L}$ of $\mathrm{O}_{2}=149 \mathrm{~g}$ of $\mathrm{KCl}$ at $\mathrm{STP}$
$\quad 33.6 \mathrm{~L}$ of $\mathrm{O}_{2}=\mathrm{x} \mathrm{g}$ of $\mathrm{KCl}$
$\therefore x=\frac{149 \times 33.6}{67.2}=74.5 \mathrm{~g}$ of $\mathrm{KCl}$
Now, $67.2 \mathrm{~L}$ of $\mathrm{O}_{2}=149 \mathrm{~g}$ of $\mathrm{KCl}$ at $\mathrm{STP}$
$\quad 33.6 \mathrm{~L}$ of $\mathrm{O}_{2}=\mathrm{x} \mathrm{g}$ of $\mathrm{KCl}$
$\therefore x=\frac{149 \times 33.6}{67.2}=74.5 \mathrm{~g}$ of $\mathrm{KCl}$
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