One gram of an alloy of aluminium and magnesium when heated with excess of dil. `HCI` forms magnesium chloride, aluminium chloride and hydrogen. The evolved hydrogen collected over mercury at `0^(0)C` has a volume of `1.2` litre at `0.92 atm` pressure. Calculate the composition of the alloy.

Explanation

To solve the problem, we need to find the composition of the alloy of aluminum (Al) and magnesium (Mg) based on the volume of hydrogen gas evolved when the alloy is reacted with dilute hydrochloric acid (HCl). Here’s a step-by-step solution:

### Step 1: Write the balanced chemical equations
1. **For magnesium:**
\[ \text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 \]
This shows that 1 mole of magnesium produces 1 mole of hydrogen.

2. **For aluminum:**
\[ 2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2 \]
This shows that 2 moles of aluminum produce 3 moles of hydrogen, or 1 mole of aluminum produces \( \frac{3}{2} \) moles of hydrogen.

### Step 2: Calculate the moles of hydrogen produced
Using the ideal gas equation \( PV = nRT \), we can find the number of moles of hydrogen gas (H₂) produced.

Given:
- Pressure (P) = 0.92 atm
- Volume (V) = 1.2 L
- R (ideal gas constant) = 0.0821 L·atm/(K·mol)
- Temperature (T) = 0°C = 273 K

Using the formula:
\[ n = \frac{PV}{RT} \]
Substituting the values:
\[ n = \frac{0.92 \times 1.2}{0.0821 \times 273} = \frac{1.104}{22.414} \approx 0.0493 \text{ moles of } H_2 \]

### Step 3: Set up the equations for the alloy
Let \( x \) be the mass of magnesium in grams, then the mass of aluminum will be \( 1 - x \) grams (since the total mass of the alloy is 1 gram).

- Moles of magnesium:
\[ \text{Moles of Mg} = \frac{x}{24.3} \]

- Moles of aluminum:
\[ \text{Moles of Al} = \frac{1 - x}{27} \]

### Step 4: Relate moles of metals to moles of hydrogen
From the balanced equations:
- Moles of hydrogen from magnesium:
\[ \text{Moles of H}_2 \text{ from Mg} = \frac{x}{24.3} \]
- Moles of hydrogen from aluminum:
\[ \text{Moles of H}_2 \text{ from Al} = \frac{3}{2} \times \frac{1 - x}{27} \]

### Step 5: Total moles of hydrogen
The total moles of hydrogen produced from both metals is:
\[ \frac{x}{24.3} + \frac{3}{2} \times \frac{1 - x}{27} = 0.0493 \]

### Step 6: Solve for \( x \)
Now we can set up the equation:
\[ \frac{x}{24.3} + \frac{3(1 - x)}{54} = 0.0493 \]

Multiplying through by the least common multiple (LCM) to eliminate the denominators:
\[ 54x + 3(1 - x) \cdot 24.3 = 0.0493 \cdot 24.3 \cdot 54 \]

Calculating the right side:
\[ 0.0493 \cdot 24.3 \cdot 54 \approx 62.0 \]

Now, simplifying the left side:
\[ 54x + 72.9 - 3x = 62.0 \]
\[ 51x + 72.9 = 62.0 \]
\[ 51x = 62.0 - 72.9 \]
\[ 51x = -10.9 \]
\[ x \approx 0.454 \text{ grams of magnesium} \]

### Step 7: Calculate the mass of aluminum
The mass of aluminum is:
\[ 1 - x = 1 - 0.454 = 0.546 \text{ grams} \]

### Step 8: Conclusion
The composition of the alloy is:
- Magnesium: \( 0.454 \text{ grams} \)
- Aluminum: \( 0.546 \text{ grams} \)