The amount of zinc required to produce 224 ml of `H_(2)` at NTP on treatment with dilute `H_(2)SO_(4)` solution will be

Options

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

Explanation

To solve the problem of how much zinc is required to produce 224 mL of hydrogen gas (H₂) at Normal Temperature and Pressure (NTP) when reacting with dilute sulfuric acid (H₂SO₄), we will follow these steps:

### Step 1: Understand the Reaction
The reaction between zinc (Zn) and dilute sulfuric acid (H₂SO₄) can be represented by the following equation:
\[ \text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2 \]
From this equation, we see that 1 mole of zinc produces 1 mole of hydrogen gas.

### Step 2: Calculate the Number of Moles of Hydrogen Gas
At NTP, 1 mole of any gas occupies 22.4 liters (or 22400 mL). We need to find out how many moles of hydrogen gas are in 224 mL.
\[ \text{Number of moles of } H_2 = \frac{\text{Volume of } H_2}{\text{Molar volume at NTP}} = \frac{224 \text{ mL}}{22400 \text{ mL/mol}} = 0.01 \text{ moles} \]

### Step 3: Relate Moles of Zinc to Moles of Hydrogen
From the balanced chemical equation, we know that 1 mole of zinc produces 1 mole of hydrogen gas. Therefore, the number of moles of zinc required to produce 0.01 moles of hydrogen gas is also 0.01 moles.

### Step 4: Calculate the Mass of Zinc Required
The molar mass of zinc (Zn) is approximately 65 grams/mole. To find the mass of zinc needed for 0.01 moles, we use the formula:
\[ \text{Mass of Zn} = \text{Number of moles} \times \text{Molar mass} = 0.01 \text{ moles} \times 65 \text{ g/mole} = 0.65 \text{ grams} \]

### Final Answer
The amount of zinc required to produce 224 mL of hydrogen gas at NTP is **0.65 grams**.

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