`H_(2)` evolved at STP on complete reaction of 27 g of aluminium with excess of aqueous NaOH would be

Options

A. (a)22.4 litres
B. (b)44.8 litres
C. (c )67.2 litres
D. (d)33.6 litres

Explanation

To solve the problem of how much hydrogen gas (H₂) is evolved at STP from the complete reaction of 27 g of aluminum (Al) with excess aqueous sodium hydroxide (NaOH), we can follow these steps:

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

\[ 2 \text{Al} + 2 \text{NaOH} + 6 \text{H}_2\text{O} \rightarrow 2 \text{NaAl(OH)}_4 + 3 \text{H}_2 \]

From the balanced equation, we see that 2 moles of aluminum produce 3 moles of hydrogen gas.

### Step 2: Calculate the number of moles of aluminum
To find the number of moles of aluminum in 27 g, we use the formula:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

The molar mass of aluminum (Al) is approximately 27 g/mol. Therefore:

\[ \text{Number of moles of Al} = \frac{27 \text{ g}}{27 \text{ g/mol}} = 1 \text{ mol} \]

### Step 3: Determine the moles of hydrogen gas produced
From the balanced equation, we know that 2 moles of Al produce 3 moles of H₂. Therefore, for 1 mole of Al, the moles of H₂ produced can be calculated as follows:

\[ \text{Moles of H}_2 = 1 \text{ mol Al} \times \frac{3 \text{ mol H}_2}{2 \text{ mol Al}} = \frac{3}{2} \text{ mol H}_2 \]

### Step 4: Calculate the volume of hydrogen gas at STP
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Thus, the volume of \(\frac{3}{2}\) moles of H₂ is:

\[ \text{Volume of H}_2 = \frac{3}{2} \text{ mol} \times 22.4 \text{ L/mol} = 33.6 \text{ L} \]

### Final Answer
The volume of hydrogen gas evolved at STP from the complete reaction of 27 g of aluminum with excess aqueous NaOH is **33.6 liters**.

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