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Saturday, 31 October

JEE Main 2022 · ChemistryMultiple choiceSingle correctEasyCalculation

JEE Main 29 July 2022, Shift 2, Chemistry Q61

Question 61 of 90 in this shift, Chemistry question 1 of 30, Section A.

Consider the reaction 4 HNO3(l)+3 KCl(s)→Cl2(g)+NOCl(g)+2 H2O(g)+3 KNO3(s)4\ \mathrm{HNO_3(l)} + 3\ \mathrm{KCl(s)} \rightarrow \mathrm{Cl_2(g)} + \mathrm{NOCl(g)} + 2\ \mathrm{H_2O(g)} + 3\ \mathrm{KNO_3(s)} The amount of HNO3\mathrm{HNO_3} required to produce 110.0 g of KNO3\mathrm{KNO_3} is (Given: Atomic masses of H, O, N and K are 1, 16, 14 and 39, respectively.)
  1. (1)32.2 g
  2. (2)69.4 g
  3. (3)91.5 gOfficial answer
  4. (4)162.5 g

Official answer

Option 3

NTA final key.

Chapter
Some Basic Concepts of Chemistry
Topic
Stoichiometry
Idea tested
Mass-Mass Stoichiometry

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.