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Some Basic Concepts of Chemistry question

2022 · 29 Jul · Shift 2 · Q1

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Some Basic Concepts of Chemistry
2022 · 29 Jul · Shift 2 · Q1

Some Basic Concepts of Chemistry question

2022 · 29 Jul · Shift 2 · Q1

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −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.)

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

View written solutionFree

Correct answer: C

Write the balanced reaction

$4\,\mathrm{HNO_3} + 3\,\mathrm{KCl} \rightarrow \mathrm{Cl_2} + \mathrm{NOCl} + 2\,\mathrm{H_2O} + 3\,\mathrm{KNO_3}$

From the equation:

$4$ moles of $\mathrm{HNO_3}$ produce $3$ moles of $\mathrm{KNO_3}$ .

Calculate molar mass of $\mathrm{KNO_3}$

$M(\mathrm{KNO_3}) = 39 + 14 + 3(16) = 39 + 14 + 48 = 101\,\mathrm{g\,mol^{-1}}$

Calculate moles of $\mathrm{KNO_3}$ formed from $110.0\,\mathrm{g}$

$n(\mathrm{KNO_3}) = \frac{110.0}{101} \approx 1.089\,\mathrm{mol}$

Use stoichiometric ratio to find moles of $\mathrm{HNO_3}$ required

Since $3\,\text{mol } \mathrm{KNO_3} \leftrightarrow 4\,\text{mol } \mathrm{HNO_3}$

So, $n(\mathrm{HNO_3}) = \frac{4}{3} \times 1.089 \approx 1.452\,\mathrm{mol}$

Calculate molar mass of $\mathrm{HNO_3}$

$M(\mathrm{HNO_3}) = 1 + 14 + 3(16) = 63\,\mathrm{g\,mol^{-1}}$

Calculate mass of $\mathrm{HNO_3}$ required

$m(\mathrm{HNO_3}) = 1.452 \times 63 \approx 91.5\,\mathrm{g}$

Match with options

$\boxed{91.5\,\mathrm{g}}$

So the correct option is C.

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