When the same amount of zinc is treated separately with excess of sulphuric acid and excess of sodium hydroxide, the ratio of volumes of hydrogen evolved is
(a)1:1
(b)1:2
(c)2:1
(d) 9:4

Options

A. `1:1`
B. `1:2`
C. `2:1`
D. `9:4`

Answer

To solve the problem, we need to analyze the reactions that occur when zinc is treated with excess sulfuric acid and excess sodium hydroxide.

### Step 1: Reaction of Zinc with Sulfuric Acid
When zinc (Zn) reacts with sulfuric acid (H₂SO₄), the reaction can be represented as follows:

\[ \text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2 \uparrow \]

In this reaction, one mole of zinc reacts with one mole of sulfuric acid to produce one mole of zinc sulfate and one mole of hydrogen gas (H₂). Thus, for every mole of zinc, one mole of hydrogen gas is evolved.

### Step 2: Reaction of Zinc with Sodium Hydroxide
When zinc reacts with sodium hydroxide (NaOH), the reaction can be represented as:

\[ \text{Zn} + 2 \text{NaOH} \rightarrow \text{Na}_2\text{Zn(OH)}_4 + \text{H}_2 \uparrow \]

In this reaction, one mole of zinc reacts with two moles of sodium hydroxide to produce sodium zincate and one mole of hydrogen gas. Again, for every mole of zinc, one mole of hydrogen gas is evolved.

### Step 3: Comparing the Volumes of Hydrogen Evolved
From both reactions, we observe that:
- In the reaction with sulfuric acid, 1 mole of hydrogen is produced per mole of zinc.
- In the reaction with sodium hydroxide, 1 mole of hydrogen is also produced per mole of zinc.

Since the same amount of zinc is used in both reactions, the volume of hydrogen gas evolved in both cases will be the same.

### Conclusion
The ratio of the volumes of hydrogen evolved when zinc is treated with excess sulfuric acid and excess sodium hydroxide is:

\[ \text{Volume of H}_2 \text{ from H}_2\text{SO}_4 : \text{Volume of H}_2 \text{ from NaOH} = 1 : 1 \]

Thus, the correct answer is (a) 1:1.

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