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: Write the balanced chemical reaction.
Zinc reacts with dilute sulfuric acid to produce zinc sulfate and hydrogen gas. The balanced equation is:
\[ \mathrm{Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2} \] From the balanced reaction, we see that:
\[ 1 \, \text{mol Zn} \; : \; 1 \, \text{mol } \mathrm{H_2SO_4} \; : \; 1 \, \text{mol } \mathrm{H_2} \] So, first we must calculate the moles of zinc and sulfuric acid and then determine the limiting reagent.

Step 2: Calculate the moles of zinc.
Given mass of zinc \(= 20 \, \mathrm{g}\)
Atomic mass of zinc \(= 65\)
Therefore, moles of zinc:
\[ n(\mathrm{Zn}) = \frac{20}{65} = 0.3077 \, \text{mol} \] So, zinc present \(= 0.3077\) mol.

Step 3: Calculate the mass of \( \mathrm{H_2SO_4} \) solution.
Volume of solution \(= 50 \, \mathrm{ml}\)
Density of solution \(= 1.3 \, \mathrm{g/ml}\)
Hence, mass of solution:
\[ \text{Mass of solution} = \text{Volume} \times \text{Density} \] \[ = 50 \times 1.3 = 65 \, \mathrm{g} \] So, total mass of the sulfuric acid solution is \(65 \, \mathrm{g}\).

Step 4: Calculate the mass and moles of pure \( \mathrm{H_2SO_4} \).
The solution is \(50\% \, (w/w)\), which means:
\[ 50 \, \mathrm{g} \text{ of } \mathrm{H_2SO_4} \text{ is present in } 100 \, \mathrm{g} \text{ of solution} \] Therefore, mass of pure \( \mathrm{H_2SO_4} \) in \(65 \, \mathrm{g}\) solution:
\[ \text{Mass of } \mathrm{H_2SO_4} = \frac{50}{100} \times 65 = 32.5 \, \mathrm{g} \] Molar mass of \( \mathrm{H_2SO_4} \):
\[ 2(A) + 32 + 4(16) = 2 + 32 + 64 = 98 \, \mathrm{g/mol} \] So, moles of sulfuric acid:
\[ n(\mathrm{H_2SO_4}) = \frac{32.5}{98} = 0.3316 \, \text{mol} \] Thus, sulfuric acid present \(= 0.3316\) mol.

Step 5: Identify the limiting reagent.
From the reaction:
\[ \mathrm{Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2} \] the mole ratio of Zn and \( \mathrm{H_2SO_4} \) is \(1:1\).
Available moles:
\[ n(\mathrm{Zn}) = 0.3077 \, \text{mol} \] \[ n(\mathrm{H_2SO_4}) = 0.3316 \, \text{mol} \] Since zinc has fewer moles, \(\mathrm{Zn}\) is the limiting reagent. Therefore, the amount of hydrogen gas formed will depend on zinc.

Step 6: Calculate the moles of \( \mathrm{H_2} \) produced.
From the balanced reaction:
\[ 1 \, \text{mol Zn} \rightarrow 1 \, \text{mol } \mathrm{H_2} \] Hence, moles of hydrogen gas produced:
\[ n(\mathrm{H_2}) = 0.3077 \, \text{mol} \]
Step 7: Convert moles of hydrogen into volume at STP.
At STP, \(1\) mole of any gas occupies \(22.4 \, \mathrm{L}\).
Therefore, volume of hydrogen gas:
\[ V(\mathrm{H_2}) = 0.3077 \times 22.4 \] \[ = 6.892 \, \mathrm{L} \] This value is approximately \(6.9 \, \mathrm{L}\). In competitive exam calculations, based on rounding and option matching, the nearest given option is \(6.81 \, \mathrm{L}\).

Step 8: Compare with the options.
• (A) 6.81 L: Correct according to the given answer key and nearest matching value. • (B) 7.22 L: Incorrect. This is higher than the calculated hydrogen volume. • (C) 3.4 L: Incorrect. This is nearly half the required value. • (D) 1.46 L: Incorrect. This is far too small.
Step 9: Conclusion.
Thus, zinc is the limiting reagent, and the volume of hydrogen gas evolved at STP is taken as \(6.81 \, \mathrm{L}\) according to the given options.
Final Answer: 6.81 L