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 problem of determining the amount of zinc required to produce 1.12 mL of H₂ at STP when treated with dilute HCl, we can follow these steps:

### Step 1: Write the Balanced Chemical Equation
The reaction between zinc (Zn) and hydrochloric acid (HCl) can be represented as:
\[ \text{Zn} + 2 \text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \]

### Step 2: Determine the Moles of H₂ Produced
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters (or 22400 mL). We need to convert the given volume of hydrogen gas (1.12 mL) to moles.

Using the formula:
\[ \text{Number of moles of } H_2 = \frac{\text{Volume of } H_2}{\text{Molar volume at STP}} = \frac{1.12 \text{ mL}}{22400 \text{ mL/mol}} \]

Calculating the moles:
\[ \text{Number of moles of } H_2 = \frac{1.12}{22400} = 5 \times 10^{-5} \text{ moles} \]

### Step 3: Relate Moles of Zn to Moles of H₂
From the balanced equation, we see that 1 mole of Zn produces 1 mole of H₂. Therefore, the moles of Zn required will be the same as the moles of H₂ produced:
\[ \text{Number of moles of Zn} = 5 \times 10^{-5} \text{ moles} \]

### Step 4: Calculate the Mass of Zinc Required
We can use the formula for mass:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]
The molar mass of zinc (Zn) is approximately 65 g/mol. Thus, we can calculate the mass of zinc required:
\[ \text{Mass of Zn} = 5 \times 10^{-5} \text{ moles} \times 65 \text{ g/mol} = 3.25 \times 10^{-3} \text{ g} = 0.00325 \text{ g} \]

### Conclusion
The amount of zinc required to produce 1.12 mL of H₂ at STP is approximately **0.00325 g**.