MCQEasyJEE 2026Stoichiometry & Calculations

JEE Chemistry 2026 Question with Solution

In the reaction, 2Al(s)+6HCl(aq)2Al3+(aq)+6Cl(aq)+3H2(g)2\text{Al(s)} + 6\text{HCl(aq)} \rightarrow 2\text{Al}^{3+}\text{(aq)} + 6\text{Cl}^-\text{(aq)} + 3\text{H}_2\text{(g)}

  • A

    11.211.2 L H2_2(g) at STP is produced for every mole of HCl consumed.

  • B

    12.212.2 L HCl(aq) is consumed for every 66 L H2_2(g) produced.

  • C

    33.633.6 L H2_2(g) is produced regardless of temperature and pressure for every mole of Al that reacts.

  • D

    67.267.2 L H2_2(g) at STP is produced for every mole of Al that reacts.

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The balanced reaction is 2Al+6HCl3H22\text{Al} + 6\text{HCl} \rightarrow 3\text{H}_2.

Find: Which statement about the amount of H2\text{H}_2 produced is correct.

From the balanced chemical equation:

2Al+6HCl3H22\text{Al} + 6\text{HCl} \rightarrow 3\text{H}_2

Step 1: Determine mole ratios. From the equation:

6 mol HCl3 mol H26\text{ mol HCl} \rightarrow 3\text{ mol H}_2 1 mol HCl12 mol H21\text{ mol HCl} \rightarrow \frac{1}{2}\text{ mol H}_2

Step 2: Convert moles of H2\text{H}_2 to volume at STP. At STP,

1 mol gas=22.4 L1\text{ mol gas} = 22.4\text{ L}

Thus,

12 mol H2=12×22.4=11.2 L\frac{1}{2}\text{ mol H}_2 = \frac{1}{2} \times 22.4 = 11.2\text{ L}

Therefore, 11.2 L of H2 at STP11.2\text{ L of H}_2\text{ at STP} is produced for every mole of HCl consumed. The correct option is A.

Option Check

Given: Use the balanced equation and STP molar volume.

Find: Which option matches the stoichiometric ratio.

For option A, 1 mol HCl1\text{ mol HCl} gives 12 mol H2\frac{1}{2}\text{ mol H}_2, which at STP is 11.2 L11.2\text{ L}, so this is correct.

Option B is not supported by the mole ratio and also refers to volume of aqueous HCl, which is not determined from the balanced equation alone.

Option C is incorrect because gas volume depends on temperature and pressure.

Option D is incorrect because from 2 mol Al3 mol H22\text{ mol Al} \rightarrow 3\text{ mol H}_2, 1 mol Al1.5 mol H21\text{ mol Al} \rightarrow 1.5\text{ mol H}_2, so at STP the volume is 1.5×22.4=33.6 L1.5 \times 22.4 = 33.6\text{ L}, not 67.2 L67.2\text{ L}.

Therefore, the correct option is A.

Common mistakes

  • Using the coefficients directly as gas volumes is incorrect because the balanced equation gives mole ratios, not volumes under all conditions. First convert moles of H2\text{H}_2 to volume only after applying STP.

  • Assuming gas volume is independent of temperature and pressure is wrong. The value 22.4 L22.4\text{ L} per mole is valid only at STP, so statements claiming the same volume regardless of conditions are incorrect.

  • Confusing the ratio for aluminum with the ratio for HCl leads to a wrong answer. From 2 mol Al3 mol H22\text{ mol Al} \rightarrow 3\text{ mol H}_2, one mole of Al gives 1.5 mol H21.5\text{ mol H}_2, not the amount obtained from one mole of HCl.

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