JEE Main 2022 · 29 July · evening · Single correct · +4 / −1

Consider the reaction
4 HNO₃(1)+3 KCl(s) → Cl₂( g)+NOCl(g)+2 H₂ O(g)+3 KNO₃( s)
The amount of HNO₃ required to produce 110.0 g of KNO₃ is
(Given: Atomic masses of H, O, N and K are 1,16,14 and 39, respectively.)

  1. (a) 32.2 g
  2. (b) 69.4 g
  3. (c) 91.5 g
  4. (d) 162.5 g
Answer and solution

Answer (c) 91.5 g

Solution

Chemistry ma’am · 4 steps

Go grams to moles, cross over with the 4 : 3 mole ratio, then come back to grams.

  1. Moles of KNO3. Molar mass of KNO₃ is 39 + 14 + 3 × 16 = 101 g mol⁻¹. So 110.0 g is 110/101 = 1.089 mol.
  2. Use the mole ratio. The balanced equation shows 4 mol HNO₃ giving 3 mol KNO₃. So you need 4/3 mol of HNO₃ for every mole of KNO₃.

    n_(HNO₃) = 4/3 × 110/101 = 440/303 = 1.452 mol

  3. Back to grams. Molar mass of HNO₃ is 1 + 14 + 48 = 63 g mol⁻¹. Mass = 1.452 × 63 = 91.5 g.
  4. So. About 91.5 g of HNO₃ is required, which is option (c).

The idea to keep. Stoichiometry always runs through moles: grams to moles of the given, multiply by the coefficient ratio (wanted over given), then moles back to grams.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Mole Concept & Stoichiometry

Asked about 210 times in JEE Main since 2020. This question is from 2022.