Question:
What volume of hydrogen gas at STP would be liberated by action of 50 mL of $H_2SO_4$ of 50% purity (density $= 1.3 \text{ g mL}^{-1}$) on 20 g of zinc?
Given : Molar mass of H, O, S, Zn are 1, 16, 32, 65 $\text{g mol}^{-1}$ respectively.
What volume of hydrogen gas at STP would be liberated by action of 50 mL of $H_2SO_4$ of 50% purity (density $= 1.3 \text{ g mL}^{-1}$) on 20 g of zinc?
Given : Molar mass of H, O, S, Zn are 1, 16, 32, 65 $\text{g mol}^{-1}$ respectively.
Given : Molar mass of H, O, S, Zn are 1, 16, 32, 65 $\text{g mol}^{-1}$ respectively.
Updated On: Apr 12, 2026
- 5.824 L
- 7.428 L
- 6.892 L
- 8.375 L
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
This problem involves stoichiometry based on the reaction between Zinc and Sulfuric acid to produce Hydrogen gas.
We must first identify the limiting reagent by calculating the number of moles of each reactant available.
: Key Formula or Approach:
Chemical Equation: \[ \text{Zn} + \text{H}_2\text{SO}_4 \to \text{ZnSO}_4 + \text{H}_2 \uparrow \]
Mass of pure substance $= \text{Volume} \times \text{Density} \times \text{Purity} / 100$.
Number of moles $= \text{Mass} / \text{Molar Mass}$.
Volume of gas at STP $= \text{moles} \times 22.4 \text{ L mol}^{-1}$.
Step 2: Detailed Explanation:
1. Calculation of moles of Zn:
Mass of Zinc $= 20 \text{ g}$.
Molar mass of Zinc $= 65 \text{ g mol}^{-1}$.
Moles of Zinc $= \frac{20}{65} \approx 0.3077 \text{ mol}$.
2. Calculation of moles of pure H_2SO_4:
Volume of solution $= 50 \text{ mL}$.
Mass of solution $= 50 \text{ mL} \times 1.3 \text{ g mL}^{-1} = 65 \text{ g}$.
Mass of pure $\text{H}_2\text{SO}_4 = 65 \text{ g} \times \frac{50}{100} = 32.5 \text{ g}$.
Molar mass of $\text{H}_2\text{SO}_4 = (2 \times 1) + 32 + (4 \times 16) = 98 \text{ g mol}^{-1}$.
Moles of $\text{H}_2\text{SO}_4 = \frac{32.5}{98} \approx 0.3316 \text{ mol}$.
3. Limiting Reagent Determination:
From the balanced equation, 1 mole of Zn reacts with 1 mole of $\text{H}_2\text{SO}_4$.
Since we have 0.3077 moles of Zn and 0.3316 moles of $\text{H}_2\text{SO}_4$, Zinc is the limiting reagent.
4. Volume of H_2 gas at STP:
Moles of $\text{H}_2$ produced $=$ Moles of Zn used $= 0.3077 \text{ mol}$.
Volume at STP $= 0.3077 \times 22.4 \text{ L} \approx 6.89248 \text{ L}$.
Step 3: Final Answer:
The volume of hydrogen gas liberated at STP is 6.892 L.
This problem involves stoichiometry based on the reaction between Zinc and Sulfuric acid to produce Hydrogen gas.
We must first identify the limiting reagent by calculating the number of moles of each reactant available.
: Key Formula or Approach:
Chemical Equation: \[ \text{Zn} + \text{H}_2\text{SO}_4 \to \text{ZnSO}_4 + \text{H}_2 \uparrow \]
Mass of pure substance $= \text{Volume} \times \text{Density} \times \text{Purity} / 100$.
Number of moles $= \text{Mass} / \text{Molar Mass}$.
Volume of gas at STP $= \text{moles} \times 22.4 \text{ L mol}^{-1}$.
Step 2: Detailed Explanation:
1. Calculation of moles of Zn:
Mass of Zinc $= 20 \text{ g}$.
Molar mass of Zinc $= 65 \text{ g mol}^{-1}$.
Moles of Zinc $= \frac{20}{65} \approx 0.3077 \text{ mol}$.
2. Calculation of moles of pure H_2SO_4:
Volume of solution $= 50 \text{ mL}$.
Mass of solution $= 50 \text{ mL} \times 1.3 \text{ g mL}^{-1} = 65 \text{ g}$.
Mass of pure $\text{H}_2\text{SO}_4 = 65 \text{ g} \times \frac{50}{100} = 32.5 \text{ g}$.
Molar mass of $\text{H}_2\text{SO}_4 = (2 \times 1) + 32 + (4 \times 16) = 98 \text{ g mol}^{-1}$.
Moles of $\text{H}_2\text{SO}_4 = \frac{32.5}{98} \approx 0.3316 \text{ mol}$.
3. Limiting Reagent Determination:
From the balanced equation, 1 mole of Zn reacts with 1 mole of $\text{H}_2\text{SO}_4$.
Since we have 0.3077 moles of Zn and 0.3316 moles of $\text{H}_2\text{SO}_4$, Zinc is the limiting reagent.
4. Volume of H_2 gas at STP:
Moles of $\text{H}_2$ produced $=$ Moles of Zn used $= 0.3077 \text{ mol}$.
Volume at STP $= 0.3077 \times 22.4 \text{ L} \approx 6.89248 \text{ L}$.
Step 3: Final Answer:
The volume of hydrogen gas liberated at STP is 6.892 L.
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