1. Write the balanced reaction

$2Al(s) + 6HCl(aq) \rightarrow 2Al^{3+}(aq) + 6Cl^-(aq) + 3H_2(g)$

From the balanced equation:

• $6$ mol $HCl$ produce $3$ mol $H_2$
• $2$ mol $Al$ produce $3$ mol $H_2$

2. Find the mole ratio between $HCl$ and $H_2$

$6\text{ mol }HCl \rightarrow 3\text{ mol }H_2$

Dividing by $3$:

$2\text{ mol }HCl \rightarrow 1\text{ mol }H_2$

So for every $1$ mol $HCl$ consumed:

$1\text{ mol }HCl \rightarrow \frac{1}{2}\text{ mol }H_2$

At STP, $1$ mol gas occupies $22.4\,L$.

Therefore, $\frac{1}{2}$ mol $H_2$ occupies:

$\frac{1}{2}\times 22.4 = 11.2\,L$

So Option A is correct.

3. Check each option

Option A

"$11.2\,L$ $H_2$ at STP is produced for every mole $HCl$ consumed"

As shown above:

$1\text{ mol }HCl \rightarrow \frac{1}{2}\text{ mol }H_2 = 11.2\,L \text{ at STP}$

✅ Correct

Option B

"$6\,L$ $HCl(aq)$ is consumed for every $3\,L$ $H_2(g)$ produced"

This compares volume of aqueous HCl solution with volume of gas. For solutions, volume depends on concentration, so liters of $HCl(aq)$ cannot be directly obtained from stoichiometric coefficients unless concentration is specified.

Hence this statement is not generally valid.

❌ Incorrect

Option C

"$33.6\,L$ $H_2$ is produced regardless of temperature and pressure for every mole that reacts"

$33.6\,L$ corresponds to:

$\frac{33.6}{22.4} = 1.5\text{ mol }H_2$

This is the amount produced when:

$1\text{ mol }Al \rightarrow \frac{3}{2}\text{ mol }H_2$

But gas volume is not independent of temperature and pressure. $33.6\,L$ is only true at STP, not regardless of $T$ and $P$.

❌ Incorrect

Option D

"$67.2$ $H_2(g)$ at STP is produced for every mole $Al$ that reacts"

From stoichiometry:

$2\text{ mol }Al \rightarrow 3\text{ mol }H_2$

So,

$1\text{ mol }Al \rightarrow \frac{3}{2}\text{ mol }H_2$

At STP:

$\frac{3}{2}\times 22.4 = 33.6\,L$

Not $67.2\,L$.

❌ Incorrect

4. Final conclusion

The only correct option is:

$\boxed{A}$