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

To solve this question, we need to analyze the stoichiometry of the chemical reaction and apply the concept of gases at Standard Temperature and Pressure (STP).

The given chemical reaction is:

\(2\text{Al}(s) + 6\text{HCl}(aq) \rightarrow 2\text{Al}^{3+}(aq) + 6\text{Cl}^-(aq) + 3\text{H}_2(g)\)

This balanced chemical equation tells us that 6 moles of HCl produce 3 moles of \(\text{H}_2\) gas. According to stoichiometry, for every 1 mole of HCl, the production of \(\text{H}_2\) gas is:

\(\frac{3}{6} = 0.5 \text{ moles of } \text{H}_2\) per mole of \(\text{HCl}\).

Now, we calculate the volume of \(\text{H}_2\) at STP. At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Thus, 0.5 moles of \(\text{H}_2\) will occupy:

\(0.5 \text{ moles} \times 22.4 \text{ L/mole} = 11.2 \text{ L}\)

Thus, the volume of \(\text{H}_2\) produced at STP for every mole of \(\text{HCl}\) consumed is 11.2 liters.

Let's evaluate the options based on the above calculations:

1. 11.2 L H₂(g) at STP is produced for every mole of HCl consumed.
2. 12.2 L HCl(aq) is consumed for every 6 L H₂(g) produced.
3. 33.6 L H₂(g) is produced regardless of temperature and pressure for every mole of Al that reacts.
4. 67.2 L H₂(g) at STP is produced for every mole of Al that reacts.

The correct answer is therefore:

$11.2$ L H$_2$(g) at STP is produced for every mole of HCl consumed.