What volume of hydrogen gas at STP would be liberated by action of 50 mL of 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.

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q51 – JEE-MAIN 2026

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Home Chemistry Basic ChemistryQuestion #43

What volume of hydrogen gas at STP would be liberated by action of

50 mL

H_{2} {SO}_{4}

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.

Clue 1 of 3: Stoichiometry and Limiting Reagent

Identify the balanced chemical reaction and determine which reactant is the limiting reagent, as it dictates the maximum amount of product formed.

🧠 Full Solution Path

Step 1: Calculate moles of reactants✦ Active

The balanced chemical reaction is:

Zn + H_{2} {SO}_{4} \to {ZnSO}_{4} + H_{2}

Given molar masses: H = $1 g/mol$ , O = $16 g/mol$ , S = $32 g/mol$ , Zn = $65 g/mol$ . Molar mass of $H_{2} {SO}_{4} = 2 (1) + 32 + 4 (16) = 98 g/mol$ . Mass of $H_{2} {SO}_{4}$ solution $= Volume \times Density = 50 mL \times 1.3 g/mL = 65 g$ . Mass of pure $H_{2} {SO}_{4} = 65 g \times 0.50 = 32.5 g$ . Moles of $H_{2} {SO}_{4} = \frac{32.5 g}{98 g/mol} \approx 0.3316 mol$ . Moles of $Zn = \frac{20 g}{65 g/mol} \approx 0.3077 mol$ .

Step 2: Identify the limiting reagent○ Expand

From the balanced equation, 1 mole of Zn reacts with 1 mole of $H_{2} {SO}_{4}$ . We have $0.3077 mol$ of Zn and $0.3316 mol$ of $H_{2} {SO}_{4}$ . Since the available moles of Zn are less than the available moles of $H_{2} {SO}_{4}$ , Zinc (Zn) is the limiting reagent.

💡 Teacher's Secret Hint

The limiting reagent determines the maximum amount of product that can be formed.

Step 3: Calculate the volume of

H_{2}

gas liberated at STP○ Expand

According to the stoichiometry, 1 mole of Zn produces 1 mole of $H_{2}$ . Therefore, moles of $H_{2}$ produced = moles of limiting reagent (Zn) $= 0.3077 mol$ . At STP, 1 mole of any gas occupies $22.4 L$ . Volume of $H_{2} = Moles \times 22.4 L/mol = 0.3077 mol \times 22.4 L/mol \approx 6.89248 L$ .

💡 Teacher's Secret Hint

Remember to use the correct molar volume for gases at STP.

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Chemistry Question 51 – JEE-MAIN 2026

Clue 1 of 3: Stoichiometry and Limiting Reagent

🧠 Full Solution Path