Aluminium liberates dihydrogen gas with both dil.HCI and aqueous NaOH. The volume ratio of dihydrogen gas evolved from equal amounts of aluminium in these reactions is

Options

A. `1:2`
B. `2:3`
C. `1:3`
D. `1:1`

Answer

To solve the problem of determining the volume ratio of dihydrogen gas evolved when aluminum reacts with dilute hydrochloric acid (HCl) and aqueous sodium hydroxide (NaOH), we will follow these steps:

### Step 1: Write the balanced chemical equations for both reactions.

1. **Reaction with Dilute HCl:**
\[ 2 \text{Al} + 6 \text{HCl} \rightarrow 2 \text{AlCl}_3 + 3 \text{H}_2 \]
- Here, 2 moles of aluminum react with 6 moles of HCl to produce 2 moles of aluminum trichloride and 3 moles of dihydrogen gas.

2. **Reaction with Aqueous NaOH:**
\[ 2 \text{Al} + 2 \text{NaOH} + 6 \text{H}_2\text{O} \rightarrow 2 \text{NaAlO}_2 + 3 \text{H}_2 \]
- In this reaction, 2 moles of aluminum react with 2 moles of NaOH and 6 moles of water to produce 2 moles of sodium meta-aluminate and 3 moles of dihydrogen gas.

### Step 2: Determine the amount of dihydrogen gas produced in each reaction.

From both balanced equations, we can see that:
- In the reaction with dilute HCl, 3 moles of H2 are produced.
- In the reaction with aqueous NaOH, 3 moles of H2 are also produced.

### Step 3: Calculate the volume ratio of dihydrogen gas evolved.

Since the amount of dihydrogen gas produced in both reactions is the same (3 moles), the volume ratio of dihydrogen gas evolved from equal amounts of aluminum in these reactions is:
\[ \text{Volume ratio} = \frac{\text{Volume of } H_2 \text{ from HCl}}{\text{Volume of } H_2 \text{ from NaOH}} = \frac{3}{3} = 1:1 \]

### Conclusion

The volume ratio of dihydrogen gas evolved from equal amounts of aluminum in these reactions is **1:1**.

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