JEE Main 2026 · 5 Apr · Shift 2 · Chemistry Q51

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ChemistrySingle correct+4 / -1Question 51 of 75

What volume of hydrogen gas at STP would be liberated by action of 50 mL of $\ce{H2SO4}$ of 50% purity (density = 1.3 g mL $^{-1}$ ) on 20 g of zinc ?

Given : Molar mass of H, O, S, Zn are 1, 16, 32, 65 g mol $^{-1}$ respectively.

5.824 L

7.428 L

6.892 L

8.375 L

Answer and worked solution

Official answer: C

Moles of acid. $50\times1.3\times0.5=32.5$ g of $\ce{H2SO4}$ , which is $\tfrac{32.5}{98}=0.332$ mol.

Moles of zinc. $\tfrac{20}{65}=0.308$ mol.

Limiting reagent. $\ce{Zn + H2SO4 -> ZnSO4 + H2}$ is 1 : 1, so zinc limits.

Hydrogen. $0.3077\times22.4\approx6.892$ L.

(C)

Topics:Stoichiometry Mole concept

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JEE Main 2026 · 5 Apr · Shift 2, Chemistry question 51