The amount of metallic Zn (Atomic weight = 65.4) required to react with aqueous sodium hydroxide to produce 1 g of `H_2`, is

Options

A. 32.7 g
B. 98.1 g
C. 65.4 g
D. 16.3 g

Explanation

To find the amount of metallic zinc (Zn) required to produce 1 g of hydrogen gas (H₂) when reacting with sodium hydroxide (NaOH), we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between zinc and sodium hydroxide in the presence of water can be represented as follows:

\[ \text{Zn} + 2 \text{NaOH} \rightarrow \text{Na}_2\text{Zn(OH)}_4 + \text{H}_2 \]

This equation shows that 1 mole of zinc produces 1 mole of hydrogen gas.

### Step 2: Calculate the moles of hydrogen gas
We know that the molar mass of hydrogen gas (H₂) is approximately 2 g/mol. Therefore, the number of moles of hydrogen gas produced from 1 g is calculated as follows:

\[ \text{Moles of H}_2 = \frac{\text{mass of H}_2}{\text{molar mass of H}_2} = \frac{1 \text{ g}}{2 \text{ g/mol}} = 0.5 \text{ moles} \]

### Step 3: 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, if we need 0.5 moles of hydrogen gas, we will also need 0.5 moles of zinc.

### Step 4: Calculate the mass of zinc required
The atomic weight of zinc (Zn) is given as 65.4 g/mol. To find the mass of zinc required for 0.5 moles, we use the formula:

\[ \text{Mass of Zn} = \text{moles of Zn} \times \text{molar mass of Zn} = 0.5 \text{ moles} \times 65.4 \text{ g/mol} = 32.7 \text{ g} \]

### Conclusion
The amount of metallic zinc required to react with aqueous sodium hydroxide to produce 1 g of hydrogen gas is **32.7 g**.

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