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 how much hydrogen gas (H₂) is evolved when 0.9 g of aluminum (Al) reacts with dilute hydrochloric acid (HCl), we can follow these steps:

### Step 1: Write the balanced chemical equation.
The reaction between aluminum and hydrochloric acid can be represented as:
\[ 2 \text{Al} + 6 \text{HCl} \rightarrow 2 \text{AlCl}_3 + 3 \text{H}_2 \]

### Step 2: Calculate the number of moles of aluminum.
To find the number of moles of aluminum, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
Given:
- Mass of Al = 0.9 g
- Molar mass of Al = 27 g/mol

Calculating the moles:
\[ \text{Number of moles of Al} = \frac{0.9 \text{ g}}{27 \text{ g/mol}} = 0.03333 \text{ mol} \]

### Step 3: Use the stoichiometry of the reaction to find moles of H₂ produced.
From the balanced equation, we see that 2 moles of Al produce 3 moles of H₂. Therefore, the ratio of Al to H₂ is:
\[ \frac{3 \text{ moles H}_2}{2 \text{ moles Al}} \]
Using this ratio, we can calculate the moles of H₂ produced from the moles of Al:
\[ \text{Moles of H}_2 = 0.03333 \text{ mol Al} \times \frac{3 \text{ mol H}_2}{2 \text{ mol Al}} = 0.0500 \text{ mol H}_2 \]

### Step 4: Calculate the volume of H₂ at STP.
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of H₂ produced can be calculated as:
\[ \text{Volume of H}_2 = \text{moles of H}_2 \times 22.4 \text{ L/mol} \]
Calculating the volume:
\[ \text{Volume of H}_2 = 0.0500 \text{ mol} \times 22.4 \text{ L/mol} = 1.12 \text{ L} \]

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

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