Calculate the volume of hydrogen liberated at N.T.P. when `500 cm^(3)` of `0.5 N` sulphuric acid react with excess of zinc.

Explanation

To solve the problem of calculating the volume of hydrogen liberated at normal temperature and pressure (N.T.P.) when 500 cm³ of 0.5 N sulfuric acid reacts with excess zinc, we can follow these steps:

### Step 1: Calculate the equivalent mass of sulfuric acid
The equivalent mass of sulfuric acid (H₂SO₄) can be calculated using its molar mass and the number of replaceable hydrogen ions (n-factor).

- Molar mass of H₂SO₄ = 98 g/mol
- n-factor = 2 (since it can donate 2 H⁺ ions)

\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{98 \, \text{g/mol}}{2} = 49 \, \text{g/equiv} \]

### Step 2: Calculate the mass of sulfuric acid in the solution
Using the normality and volume of the sulfuric acid solution, we can find the mass of sulfuric acid.

- Normality (N) = 0.5 N
- Volume = 500 cm³ = 0.5 L (since 1000 cm³ = 1 L)

Using the formula for normality:

\[ \text{Normality} = \frac{\text{number of equivalents}}{\text{volume in L}} \]

We can rearrange this to find the number of equivalents:

\[ \text{Number of equivalents} = \text{Normality} \times \text{Volume in L} = 0.5 \, \text{N} \times 0.5 \, \text{L} = 0.25 \, \text{equiv} \]

Now, we can calculate the mass of sulfuric acid:

\[ \text{Mass of H₂SO₄} = \text{Number of equivalents} \times \text{Equivalent mass} = 0.25 \, \text{equiv} \times 49 \, \text{g/equiv} = 12.25 \, \text{g} \]

### Step 3: Calculate the volume of hydrogen gas produced
From the balanced chemical reaction:

\[ \text{Zn} + \text{H₂SO₄} \rightarrow \text{ZnSO₄} + \text{H₂} \]

We know that 98 g of H₂SO₄ produces 22,400 cm³ of H₂ at N.T.P.

Now, we can find the volume of hydrogen produced from 12.25 g of H₂SO₄:

\[ \text{Volume of H₂} = \frac{22400 \, \text{cm}^3}{98 \, \text{g}} \times 12.25 \, \text{g} \]

Calculating this gives:

\[ \text{Volume of H₂} = \frac{22400}{98} \times 12.25 = 2800 \, \text{cm}^3 \]

### Final Answer
The volume of hydrogen liberated at N.T.P. when 500 cm³ of 0.5 N sulfuric acid reacts with excess zinc is **2800 cm³**.

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