27 gm Al is heated with 49 ml of `H_2SO_4`(specific gravity =2) produces` H_2` gas. Percentage of Al reacted with `H_2SO_4` is:

Explanation

To solve the problem, we need to determine the percentage of aluminum (Al) that reacts with sulfuric acid (H₂SO₄). Here’s a step-by-step solution:

### Step 1: Write the balanced chemical equation
The reaction between aluminum and sulfuric acid can be represented as follows:

\[ 2 \text{Al} + 3 \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3 \text{H}_2 \]

### Step 2: Calculate the molar mass of reactants
- The molar mass of aluminum (Al) = 27 g/mol
- The molar mass of sulfuric acid (H₂SO₄) = 98 g/mol

### Step 3: Calculate the mass of H₂SO₄ used
Given that the specific gravity of H₂SO₄ is 2 and the volume is 49 mL, we can calculate the mass of H₂SO₄:

\[ \text{Mass of H}_2\text{SO}_4 = \text{Volume} \times \text{Specific Gravity} = 49 \, \text{mL} \times 2 \, \text{g/mL} = 98 \, \text{g} \]

### Step 4: Determine the moles of H₂SO₄
Now, we can calculate the moles of H₂SO₄ used:

\[ \text{Moles of H}_2\text{SO}_4 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{98 \, \text{g}}{98 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 5: Determine the moles of aluminum that can react
From the balanced equation, we see that 3 moles of H₂SO₄ react with 2 moles of Al. Therefore, 1 mole of H₂SO₄ will react with:

\[ \text{Moles of Al} = \frac{2}{3} \times \text{Moles of H}_2\text{SO}_4 = \frac{2}{3} \times 1 = \frac{2}{3} \, \text{mol} \]

### Step 6: Calculate the mass of aluminum that reacts
Now, we can calculate the mass of aluminum that reacts:

\[ \text{Mass of Al reacted} = \text{Moles of Al} \times \text{Molar Mass of Al} = \frac{2}{3} \, \text{mol} \times 27 \, \text{g/mol} = 18 \, \text{g} \]

### Step 7: Calculate the percentage of aluminum reacted
Finally, we can find the percentage of aluminum that reacted with sulfuric acid:

\[ \text{Percentage of Al reacted} = \left( \frac{\text{Mass of Al reacted}}{\text{Initial mass of Al}} \right) \times 100 = \left( \frac{18 \, \text{g}}{27 \, \text{g}} \right) \times 100 = 66.67\% \]

### Final Answer:
The percentage of aluminum that reacted with sulfuric acid is approximately **66.67%**.

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