Each of the following metal on reaction with dilute hydrochloric acid produces `H_(2)` (g) . In which of the following reaction of metal, the molar ratio of the metal used and hydrogen produced is `2 : 3` ?

Options

A. Fe
B. Al
C. Si
D. Zn

Answer

To determine which metal reacts with dilute hydrochloric acid to produce hydrogen gas in a molar ratio of 2:3, we can analyze the reactions of different metals with hydrochloric acid. Let's go through the reactions step by step.

### Step 1: Identify the Reactions

1. **Iron (Fe) with Hydrochloric Acid (HCl)**:
\[ \text{Fe} + 2 \text{HCl} \rightarrow \text{FeCl}_2 + \text{H}_2 \]
- Here, 1 mole of iron reacts with 2 moles of HCl to produce 1 mole of hydrogen gas.

2. **Aluminum (Al) with Hydrochloric Acid (HCl)**:
\[ 2 \text{Al} + 6 \text{HCl} \rightarrow 2 \text{AlCl}_3 + 3 \text{H}_2 \]
- In this reaction, 2 moles of aluminum react with 6 moles of HCl to produce 3 moles of hydrogen gas.

### Step 2: Calculate the Molar Ratios

1. **For Iron**:
- Molar ratio of Fe to H₂:
\[ \text{Fe} : \text{H}_2 = 1 : 1 \]

2. **For Aluminum**:
- Molar ratio of Al to H₂:
\[ \text{Al} : \text{H}_2 = 2 : 3 \]

### Step 3: Conclusion

From the calculations, we find that:
- The reaction of iron gives a molar ratio of 1:1.
- The reaction of aluminum gives a molar ratio of 2:3.

Thus, the metal that produces hydrogen gas in a molar ratio of 2:3 when reacting with dilute hydrochloric acid is **Aluminum (Al)**.

### Final Answer:
The answer is **Aluminum (Al)**.

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