The amount of zinc required to produce 1.12 ml of `H_(2)` at STP on treatment with dilute HCI will be

Options

A. 65 g
B. 0.065 g
C. `32.5 xx 10^(-4) g`
D. 6.5 g

Answer

To solve the question of how much zinc is required to produce 1.12 ml of hydrogen gas (H₂) at STP when treated with dilute hydrochloric acid (HCl), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between zinc (Zn) and hydrochloric acid (HCl) can be represented as:
\[ \text{Zn} + 2 \text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \]
From the balanced equation, we can see that 1 mole of zinc produces 1 mole of hydrogen gas.

2. **Convert Volume of H₂ to Moles**:
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. First, we need to convert the volume of hydrogen gas from milliliters to liters:
\[ 1.12 \, \text{ml} = 1.12 \times 10^{-3} \, \text{L} \]
Now, we can calculate the number of moles of hydrogen gas produced:
\[ \text{Number of moles of } H_2 = \frac{\text{Volume (L)}}{22.4 \, \text{L/mol}} = \frac{1.12 \times 10^{-3}}{22.4} \approx 0.00005 \, \text{mol} \]

3. **Determine Moles of Zinc Required**:
From the stoichiometry of the reaction, we know that 1 mole of zinc produces 1 mole of hydrogen gas. Therefore, the moles of zinc required will be the same as the moles of hydrogen produced:
\[ \text{Moles of Zn required} = 0.00005 \, \text{mol} \]

4. **Calculate the Mass of Zinc**:
To find the mass of zinc required, we use the formula:
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \]
The molar mass of zinc (Zn) is approximately 65 g/mol. Thus, the mass of zinc required is:
\[ \text{Mass of Zn} = 0.00005 \, \text{mol} \times 65 \, \text{g/mol} = 0.00325 \, \text{g} = 3.25 \times 10^{-3} \, \text{g} \]

5. **Final Result**:
The amount of zinc required to produce 1.12 ml of hydrogen gas at STP is approximately:
\[ 3.25 \times 10^{-3} \, \text{g} \text{ or } 0.00325 \, \text{g} \]

### Summary:
The amount of zinc required to produce 1.12 ml of H₂ at STP is approximately **0.00325 g**.