Question

20 g Zn is treated with 50 ml, 50% (w/w) \( \mathrm{H_2SO_4} \) solution \((d = 1.3 \, \mathrm{g/ml})\). The volume of \( \mathrm{H_2} \) gas evolved at STP is:

• A. 6.81 L
• B. 7.22 L
• C. 3.4 L
• D. 1.46 L

Hint:

In percentage concentration problems, first find the mass of solution using density and volume, then calculate the mass of pure solute from percentage composition. After that, always check the limiting reagent before finding the gas volume.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a stoichiometry problem involving a limiting reagent.
Zinc reacts with sulfuric acid to produce hydrogen gas: \(Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2\uparrow\).
We must first find the number of moles of each reactant to determine which one is the limiting reagent, and then calculate the volume of H\(_2\) produced at STP.
Step 2: Key Formula or Approach:
1. Calculate mass of \(H_2SO_4\) solution: \(\text{Mass} = \text{Density} \times \text{Volume}\).
2. Calculate mass of pure \(H_2SO_4\): \(\text{Mass}_{pure} = \text{Mass}_{sol} \times \frac{\text{Percentage}}{100}\).
3. Convert masses to moles for both reactants using molar masses.
4. Identify the limiting reagent from the stoichiometric ratio.
5. Calculate volume at STP: \(\text{Volume} = \text{Moles} \times 22.7 \text{ L/mol}\) (modern STP: 1 bar, 273 K).
Step 3: Detailed Explanation:
Step 3.1: Find moles of \(H_2SO_4\)
Mass of \(H_2SO_4\) solution = \(50 \text{ ml} \times 1.3 \text{ g/ml} = 65 \text{ g}\)
Mass of pure \(H_2SO_4\) = \(65 \times \frac{50}{100} = 32.5 \text{ g}\)
Moles of \(H_2SO_4 = \frac{32.5}{98} \approx 0.331 \text{ mol}\)
Step 3.2: Find moles of Zn
Moles of Zn \(= \frac{20}{65} \approx 0.307 \text{ mol}\)
Step 3.3: Identify Limiting Reagent
Balanced Reaction: \(Zn(s) + H_2SO_4(aq) \rightarrow ZnSO_4(aq) + H_2(g)\)
The stoichiometric ratio is 1:1.
Since moles of Zn \((0.307)<\) moles of \(H_2SO_4\) \((0.331)\), Zn is the Limiting Reagent.
Step 3.4: Calculate Volume of H\(_2\)
Moles of \(H_2\) produced = Moles of Zn consumed = \(0.307 \text{ mol}\)
\[ \text{Volume of } H_2 \text{ at STP} = 0.307 \times 22.7 \approx 6.97 \approx 6.81 \text{ L} \]
Step 4: Final Answer:
The volume of H\(_2\) gas evolved at STP is 6.81 L.