Question

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

• A. $11.2$ L H$_2$(g) at STP is produced for every mole of HCl consumed.
• B. $12.2$ L HCl$(aq)$ is consumed for every $6$ L H$_2$(g)$ produced.
• C. $33.6$ L H$_2$(g) is produced regardless of temperature and pressure for every mole of Al that reacts.
• D. $67.2$ L H$_2$(g) at STP is produced for every mole of Al that reacts.

Hint:

Always use mole ratios from the balanced equation before converting gases to volume using STP conditions.

Answer 1

The Correct Option is A

From the balanced chemical equation: \[ 2\text{Al} + 6\text{HCl} \rightarrow 3\text{H}_2 \] Step 1: Determine mole ratios.
From the equation: \[ 6\text{ mol HCl} \rightarrow 3\text{ mol H}_2 \] \[ 1\text{ mol HCl} \rightarrow \frac{1}{2}\text{ mol H}_2 \] Step 2: Convert moles of H$_2$ to volume at STP.
At STP, \[ 1\text{ mol gas} = 22.4\text{ L} \] Thus, \[ \frac{1}{2}\text{ mol H}_2 = \frac{1}{2} \times 22.4 = 11.2\text{ L} \] Final Answer: $\boxed{11.2\text{ L of H}_2 \text{ at STP}}$