The amount of zinc required to produce 224 ml of `H_(2)` at STP on treatment with dilute `H_(2)SO_(4)` will be (Zn = 65)

Options

A. 65 g
B. 0.065 g
C. 0.65 g
D. 6.5 g

Answer

To find the amount of zinc required to produce 224 ml of hydrogen gas (H₂) at standard temperature and pressure (STP) when treated with dilute sulfuric acid (H₂SO₄), we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between zinc (Zn) and dilute sulfuric acid (H₂SO₄) can be represented as:
\[ \text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2 \]

From the balanced equation, we can see that 1 mole of zinc produces 1 mole of hydrogen gas.

### Step 2: Calculate the number of moles of hydrogen gas produced
At STP, 1 mole of any gas occupies 22.4 liters (or 22400 ml). Therefore, to find the number of moles of hydrogen gas produced from 224 ml, we can use the formula:
\[ \text{Number of moles} = \frac{\text{Volume (ml)}}{22400 \text{ ml}} \]

Substituting the values:
\[ \text{Number of moles of } H_2 = \frac{224 \text{ ml}}{22400 \text{ ml}} = 0.01 \text{ moles} \]

### Step 3: Relate moles of zinc to moles of hydrogen gas
From the balanced equation, we know that 1 mole of zinc produces 1 mole of hydrogen gas. Therefore, the number of moles of zinc required will also be 0.01 moles.

### Step 4: Calculate the mass of zinc required
To find the mass of zinc required, we can use the formula:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]

Given that the molar mass of zinc (Zn) is 65 g/mol, we can substitute the values:
\[ \text{Mass of Zn} = 0.01 \text{ moles} \times 65 \text{ g/mol} = 0.65 \text{ g} \]

### Conclusion
The amount of zinc required to produce 224 ml of hydrogen gas at STP is **0.65 g**.

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