0.9 g Al reacts with dil. HCl to give `H_(2)`. The volume of `H_(2)` evolved at STP is (Atomic weight of Al = 27)

Options

A. 1.12 litre
B. 2.24 litre
C. 3.33 litre
D. 4.44 litre

Answer

To solve the problem of calculating the volume of hydrogen gas evolved when 0.9 g of aluminum reacts with dilute HCl, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction of aluminum with hydrochloric acid can be represented as:
\[ 2 \text{Al} + 6 \text{HCl} \rightarrow 2 \text{AlCl}_3 + 3 \text{H}_2 \]
This shows that 2 moles of aluminum produce 3 moles of hydrogen gas.

2. **Calculate Moles of Aluminum:**
To find the number of moles of aluminum, we use the formula:
\[ \text{Moles of Al} = \frac{\text{Mass of Al}}{\text{Molar Mass of Al}} \]
Given that the mass of aluminum is 0.9 g and the molar mass of aluminum (Al) is 27 g/mol:
\[ \text{Moles of Al} = \frac{0.9 \, \text{g}}{27 \, \text{g/mol}} = \frac{0.9}{27} = \frac{1}{30} \, \text{mol} \]

3. **Determine Moles of Hydrogen Gas Produced:**
From the balanced equation, we see that 2 moles of aluminum produce 3 moles of hydrogen gas. Therefore, we can set up a ratio:
\[ \frac{2 \, \text{mol Al}}{3 \, \text{mol H}_2} = \frac{1/30 \, \text{mol Al}}{x \, \text{mol H}_2} \]
Solving for \(x\):
\[ x = \frac{3}{2} \times \frac{1}{30} = \frac{3}{60} = \frac{1}{20} \, \text{mol H}_2 \]

4. **Calculate the Volume of Hydrogen Gas at STP:**
At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 liters. Therefore, the volume of hydrogen gas produced can be calculated as:
\[ \text{Volume of H}_2 = \text{Moles of H}_2 \times 22.4 \, \text{L/mol} \]
Substituting the moles of hydrogen gas:
\[ \text{Volume of H}_2 = \frac{1}{20} \, \text{mol} \times 22.4 \, \text{L/mol} = \frac{22.4}{20} = 1.12 \, \text{L} \]

### Final Answer:
The volume of hydrogen gas evolved at STP is **1.12 liters**.

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