Consider the reaction 4 HNO _ 3 (l) + 3KCl(s) → Cl _ 2 (g) + NOCl(g) + 2 H _ 2 O(g) + 3 KNO _ 3 (s) The amount of HNO _ 3 required to produce 110 . 0g of KNO _ 3 is (Given : Atomic masses of H, O, N and K are 1, 16, 14

Correct answer: C. 91 . 5g 4 HNO _ 3 (l) + 3KCl(s) → Cl _ 2 (g) + NOCl(g) + 2 H _ 2 O(g) + 3 KNO _ 3 (g) HNO _ 3 KNO _ 3 WeightXgm110gmmoles(x)/(63)(110)/(101) As per the stoichiometric equation 4molesof HNO _ 3 → 3molesof KNO _ 3 1 → (3)/(4) (x)/(63) → (3)/(4) × (x)/(63) = (110)/(101) x = (110 × 63 × 4)/(101 × 3) = 91 . 5gm

JEE Main 202229 JulyEvening Shift

Exam: JEE Main
Year: 2022
Date: 29 July
Shift: Evening Shift
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

JEE Main 2022 · 29 July · Evening Shift · Chemistry · Some Basic Concepts of Chemistry

Consider the reaction 4 HNO _ 3 (l) + 3KCl(s) → Cl _ 2 (g) +...

Consider the reaction

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

The amount of ${HNO}_{3}$ required to produce $110 . 0g$ of ${KNO}_{3}$ is

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

Options

A $32 . 2g$
B $69 . 4g$
C $91 . 5g$
D $162 . 5g$

Correct answer

C. $91 . 5g$

Step-by-step solution

$4{HNO}_{3}\left(l\right) + 3KCl\left(s\right) \to {Cl}_{2}\left(g\right) + NOCl\left(g\right) + 2{H}_{2}O\left(g\right) + 3{KNO}_{3}\left(g\right)$
${HNO}_{3}{KNO}_{3}WeightXgm110gmmoles\dfrac{x}{63}\dfrac{110}{101}$
As per the stoichiometric equation $4molesof{HNO}_{3} \to 3molesof{KNO}_{3}$
$1 \to \dfrac{3}{4}$
$\dfrac{x}{63} \to \dfrac{3}{4} \times \dfrac{x}{63} = \dfrac{110}{101}$
$x = \dfrac{110 \times 63 \times 4}{101 \times 3} = 91 . 5gm$

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