JEE Challenger
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ChemMedMain 2026 • 5th Apr - Shift 2

Volume of Hydrogen Liberated from Sulfuric Acid and Zinc Reaction

What volume of hydrogen gas at STP would be liberated by action of 50 mL50\text{ mL} of H2SO4\text{H}_2\text{SO}_4 of 50%50\% purity (density =1.3 g mL−1= 1.3\text{ g mL}^{-1}) on 20 g20\text{ g} of zinc ? Given : Molar mass of H,O,S,Zn\text{H}, \text{O}, \text{S}, \text{Zn} are 1,16,32,65 g mol−11, 16, 32, 65\text{ g mol}^{-1} respectively.

Options

A

5.824 L5.824\text{ L}

B

7.428 L7.428\text{ L}

C

6.892 L6.892\text{ L}

Correct
D

8.375 L8.375\text{ L}

Topics & Concepts

Some Basic Concepts of ChemistryMole ConceptStoichiometry

Step-by-Step Solution

To determine the volume of hydrogen gas liberated at STP, we first write the balanced chemical equation for the reaction:

Zn (s)+H2SO4 (aq)→ZnSO4 (aq)+H2 (g)\text{Zn (s)} + \text{H}_2\text{SO}_4\text{ (aq)} \rightarrow \text{ZnSO}_4\text{ (aq)} + \text{H}_2\text{ (g)}

Step 1: Calculate the amount of pure H2SO4\text{H}_2\text{SO}_4 present

  • Mass of the sulfuric acid solution: Mass of solution=Volume×Density=50 mL×1.3 g mL−1=65 g\text{Mass of solution} = \text{Volume} \times \text{Density} = 50\text{ mL} \times 1.3\text{ g mL}^{-1} = 65\text{ g}

  • Mass of pure H2SO4\text{H}_2\text{SO}_4 (50% purity): Mass of pure H2SO4=65 g×50100=32.5 g\text{Mass of pure }\text{H}_2\text{SO}_4 = 65\text{ g} \times \frac{50}{100} = 32.5\text{ g}

  • Molar mass of H2SO4\text{H}_2\text{SO}_4: MH2SO4=2(1)+32+4(16)=98 g mol−1M_{\text{H}_2\text{SO}_4} = 2(1) + 32 + 4(16) = 98\text{ g mol}^{-1}

  • Number of moles of H2SO4\text{H}_2\text{SO}_4: nH2SO4=32.5 g98 g mol−1≈0.3316 moln_{\text{H}_2\text{SO}_4} = \frac{32.5\text{ g}}{98\text{ g mol}^{-1}} \approx 0.3316\text{ mol}

Step 2: Calculate the number of moles of Zn\text{Zn}

  • Molar mass of Zn=65 g mol−1\text{Zn} = 65\text{ g mol}^{-1}
  • Number of moles of Zn\text{Zn}: nZn=20 g65 g mol−1=413 mol≈0.3077 moln_{\text{Zn}} = \frac{20\text{ g}}{65\text{ g mol}^{-1}} = \frac{4}{13}\text{ mol} \approx 0.3077\text{ mol}

Step 3: Determine the Limiting Reagent

From the balanced chemical equation, 1 mole of Zn\text{Zn} reacts with 1 mole of H2SO4\text{H}_2\text{SO}_4.

Comparing the available moles: nZn=0.3077 mol<nH2SO4=0.3316 moln_{\text{Zn}} = 0.3077\text{ mol} < n_{\text{H}_2\text{SO}_4} = 0.3316\text{ mol}

Therefore, zinc (Zn\text{Zn}) is the limiting reagent.

Step 4: Calculate the volume of H2\text{H}_2 liberated at STP

According to the reaction stoichiometry, 1 mole of Zn\text{Zn} produces 1 mole of H2\text{H}_2 gas:

nH2=nZn=413 moln_{\text{H}_2} = n_{\text{Zn}} = \frac{4}{13}\text{ mol}

At STP, 1 mole of an ideal gas occupies 22.4 L22.4\text{ L}:

Volume of H2=413 mol×22.4 L mol−1=89.613 L≈6.8923 L\text{Volume of }\text{H}_2 = \frac{4}{13}\text{ mol} \times 22.4\text{ L mol}^{-1} = \frac{89.6}{13}\text{ L} \approx 6.8923\text{ L}

Thus, the volume of hydrogen gas liberated at STP is approximately 6.892 L6.892\text{ L}.

Correct Option: C

Volume of Hydrogen Liberated from Sulfuric Acid and Zinc Reaction | Chemistry PYQ Solution - JEE Challenger