In the reaction:
`2Al_((s))+6HCl_((aq.))rarr2Al_((aq.))^(3+)+6Cl_((aq.))^(-)+3H_(2(g))`

Options

A. `6 "litre" HCl_((aq.))` is consumed for every `3 LH_(2(g))` produced
B. `33.6 "litre" H_(2(g))` is produced regardless of temperature and pressure for every mole `Al` that react
C. `67.2 "litre" H_(2(g))` at `STP` is produced for every mole `Al` that reacts
D. `11.2 "litre" H_(2(g))` at `STP` is produced for every mole `HCl_((aq.))` consumed

Explanation

To solve the problem, we need to analyze the given reaction and the statements provided. The reaction is:

\[ 2 \text{Al}_{(s)} + 6 \text{HCl}_{(aq)} \rightarrow 2 \text{Al}^{3+}_{(aq)} + 6 \text{Cl}^-_{(aq)} + 3 \text{H}_2_{(g)} \]

### Step-by-Step Solution:

1. **Identify the Stoichiometry of the Reaction:**
- From the balanced equation, we can see that 2 moles of Al react with 6 moles of HCl to produce 3 moles of H₂ gas.

2. **Calculate the Volume of Hydrogen Produced:**
- At Standard Temperature and Pressure (STP), 1 mole of gas occupies 22.4 liters.
- Therefore, 3 moles of H₂ will occupy:
\[ 3 \text{ moles} \times 22.4 \text{ L/mole} = 67.2 \text{ L} \]

3. **Analyze Each Statement:**
- **Statement 1:** "6L of HCl is consumed for 3L of hydrogen produced."
- This statement cannot be verified without specific conditions of temperature and pressure, so it is incorrect.

- **Statement 2:** "33.6L of H₂ is produced regardless of temperature and pressure for every mole of aluminium."
- From the stoichiometry, 2 moles of Al produce 3 moles of H₂. Therefore, for 1 mole of Al:
\[ \frac{3 \text{ moles H}_2}{2 \text{ moles Al}} = 1.5 \text{ moles H}_2 \]
- At STP, this is:
\[ 1.5 \text{ moles} \times 22.4 \text{ L/mole} = 33.6 \text{ L} \]
- This statement is also incorrect as it lacks the conditions of STP.

- **Statement 3:** "67.2L of hydrogen at STP is produced for every mole of aluminium."
- This is incorrect because we calculated that 1 mole of Al produces 33.6 L of H₂, not 67.2 L.

- **Statement 4:** "11.2L hydrogen at STP is produced for every mole of HCl."
- From the reaction, 6 moles of HCl produce 3 moles of H₂. Therefore, for 1 mole of HCl:
\[ \frac{3 \text{ moles H}_2}{6 \text{ moles HCl}} = 0.5 \text{ moles H}_2 \]
- At STP, this is:
\[ 0.5 \text{ moles} \times 22.4 \text{ L/mole} = 11.2 \text{ L} \]
- This statement is correct.

### Conclusion:
- The only correct statement is **Statement 4**: "11.2L hydrogen at STP is produced for every mole of HCl."