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 solve the problem of how much zinc is 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:
\[ \text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2 + \text{ZnSO}_4 \]

### Step 2: Determine the volume of hydrogen gas produced
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters (or 22400 ml). The problem states that we need to produce 224 ml of hydrogen gas.

### Step 3: Calculate the number of moles of hydrogen gas
Using the volume of hydrogen gas produced, we can calculate the number of moles of H₂:
\[ \text{Number of moles of H}_2 = \frac{\text{Volume of H}_2}{\text{Molar volume at STP}} = \frac{224 \text{ ml}}{22400 \text{ ml/mol}} = 0.01 \text{ moles} \]

### Step 4: Relate moles of zinc to moles of hydrogen
From the balanced equation, we see that 1 mole of zinc produces 1 mole of hydrogen gas. Therefore, the number of moles of zinc required is also 0.01 moles.

### Step 5: Calculate the mass of zinc required
Now, we can calculate the mass of zinc needed using its molar mass:
\[ \text{Mass of Zn} = \text{Number of moles of Zn} \times \text{Molar mass of Zn} = 0.01 \text{ moles} \times 65 \text{ g/mol} = 0.65 \text{ grams} \]

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

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