JEE MAIN - Chemistry (2022 - 29th July Evening Shift - No. 1)
Consider the reaction
$$4 \mathrm{HNO}_{3}(1)+3 \mathrm{KCl}(\mathrm{s}) \rightarrow \mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{NOCl}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{KNO}_{3}(\mathrm{~s})$$
The amount of $$\mathrm{HNO}_{3}$$ required to produce $$110.0 \mathrm{~g}$$ of $$\mathrm{KNO}_{3}$$ is
(Given: Atomic masses of $$\mathrm{H}, \mathrm{O}, \mathrm{N}$$ and $$\mathrm{K}$$ are $$1,16,14$$ and 39, respectively.)
32.2 g
69.4 g
91.5 g
162.5 g
Explanation
$$
\begin{array}{r}
4 \mathrm{HNO}_{3}(\mathrm{l})+3 \mathrm{KCl}(\mathrm{s}) \longrightarrow \mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{NOCl}(\mathrm{g})+ \\
2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{KNO}_{3}(\mathrm{~s})
\end{array}
$$
$\because 110 \mathrm{~g}$ of $\mathrm{KNO}_{3} \Rightarrow$ moles of $\mathrm{KNO}_{3}=\frac{110}{101}$
$$ =1.089 \mathrm{~mol} $$
As, 4 mole of $\mathrm{HNO}_{3}$ produces $3 \mathrm{~mol}$ of $\mathrm{KNO}_{3}$.
Hence, the moles of $\mathrm{HNO}_{3}$ required to produce
$1.089$ moles of $\mathrm{KNO}_{3}=\frac{4}{3} \times 1.089=1.452 \mathrm{~mol}$
Hence, mass of $\mathrm{HNO}_{3}$ required is $1.452 \times 63$
$$ =91.5 \mathrm{~g} \text { } $$
$\because 110 \mathrm{~g}$ of $\mathrm{KNO}_{3} \Rightarrow$ moles of $\mathrm{KNO}_{3}=\frac{110}{101}$
$$ =1.089 \mathrm{~mol} $$
As, 4 mole of $\mathrm{HNO}_{3}$ produces $3 \mathrm{~mol}$ of $\mathrm{KNO}_{3}$.
Hence, the moles of $\mathrm{HNO}_{3}$ required to produce
$1.089$ moles of $\mathrm{KNO}_{3}=\frac{4}{3} \times 1.089=1.452 \mathrm{~mol}$
Hence, mass of $\mathrm{HNO}_{3}$ required is $1.452 \times 63$
$$ =91.5 \mathrm{~g} \text { } $$
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