Consider the reaction $4 \mathrm{HNO}_{3}(l)+3 \mathrm{KCl}(\mathrm{s}) \rightarrow \mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{NOCl}(\mathrm{g})+ 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{KNO}_{3}(\mathrm{~s})$ The amount of $\mathrm{HNO}_{3}$ required to produce 110.0 g of $\mathrm{KNO}_{3}$ is : (Given : Atomic masses of $\mathrm{H}, \mathrm{O}, \mathrm{N}$ and K are 1, 16,14 and 39 , respectively.)

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Some Basic Concepts of Chemistry

Consider the reaction

4 HNO_{3} (l) + 3 KCl (s) \to Cl_{2} (g) + NOCl (g) + 2 H_{2} O (g) + 3 KNO_{3} (s)

The amount of

HNO_{3}

required to produce 110.0 g of

KNO_{3}

is :
(Given : Atomic masses of

H, O, N

and K are 1, 16,14 and 39 , respectively.)

Answer: (C)

From the balanced equation, the stoichiometric ratio is 4 mol

HNO_{3}

: 3 mol

KNO_{3}

.

Molar mass of

KNO_{3}

= 39 + 14 + 3(16) = 101 g/mol.
Moles of

KNO_{3}

= 110/101 mol.

Using stoichiometry: moles of

HNO_{3}

= (4/3) × (110/101) mol.

Molar mass of

HNO_{3}

= 1 + 14 + 3(16) = 63 g/mol.

Mass of

HNO_{3}

= (4/3) × (110/101) × 63 = (4 × 110 × 63)/(3 × 101) = 91.5 g.

Answer: (C) 91.5 g

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Single correct option