JEE Main 2022 — Chemistry Physical Chemistry

2022mcqeasy

Consider the reaction

$4{\mathrm{HNO}}_{3}(l)+3\mathrm{KCl}(s)\rightarrow {\mathrm{Cl}}_{2}(g)+\mathrm{NOCl}(g)+2{H}_{2}O(g)+3{\mathrm{KNO}}_{3}(s)$

The amount of ${\mathrm{HNO}}_{3}$ required to produce $110.0g$ of ${\mathrm{KNO}}_{3}$ is

(Given : Atomic masses of $H,O,N$ and $K$ are $1,16,14$ and $39$ , respectively.)

Official previous-year question

Held on 29 Jul 2022 · Verified 6 Jul 2026.

Options

A
$32.2g$
B
$69.4g$
C
$91.5g$
D
$162.5g$

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Solution

$4{\mathrm{HNO}}_{3}(l)+3\mathrm{KCl}(s)\rightarrow {\mathrm{Cl}}_{2}(g)+\mathrm{NOCl}(g)+2{H}_{2}O(g)+3{\mathrm{KNO}}_{3}(g)$

$\begin{matrix} \\ \end{matrix}$ ${\mathrm{HNO}}_{3}{\mathrm{KNO}}_{3} \mathrm{Weight}\mathrm{Xgm}110\mathrm{gm} \mathrm{moles}\frac{x}{63}\frac{110}{101}$

As per the stoichiometric equation $4\mathrm{moles}\mathrm{of}{\mathrm{HNO}}_{3}\rightarrow 3\mathrm{moles}\mathrm{of}{\mathrm{KNO}}_{3}$

$1\rightarrow \frac{3}{4}$

$\frac{x}{63}\rightarrow \frac{3}{4}\times \frac{x}{63}=\frac{110}{101}$

$x=\frac{110\times 63\times 4}{101\times 3}=91.5\mathrm{gm}$

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Solution

$4{\mathrm{HNO}}_{3}(l)+3\mathrm{KCl}(s)\rightarrow {\mathrm{Cl}}_{2}(g)+\mathrm{NOCl}(g)+2{H}_{2}O(g)+3{\mathrm{KNO}}_{3}(g)$

$\begin{matrix} \\ \end{matrix}$ ${\mathrm{HNO}}_{3}{\mathrm{KNO}}_{3} \mathrm{Weight}\mathrm{Xgm}110\mathrm{gm} \mathrm{moles}\frac{x}{63}\frac{110}{101}$

As per the stoichiometric equation $4\mathrm{moles}\mathrm{of}{\mathrm{HNO}}_{3}\rightarrow 3\mathrm{moles}\mathrm{of}{\mathrm{KNO}}_{3}$

$1\rightarrow \frac{3}{4}$

$\frac{x}{63}\rightarrow \frac{3}{4}\times \frac{x}{63}=\frac{110}{101}$

$x=\frac{110\times 63\times 4}{101\times 3}=91.5\mathrm{gm}$

JEE Main 2022 — Chemistry Physical Chemistry

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JEE Main 2022 — Chemistry Physical Chemistry

Options

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Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry