With the help of the balanced chemical equation
`2Al(s)+3H_(2)SO_(4)(aq) to Al_(2)(SO_(4))_(3)(aq)+3H_(2)(g)`
Answer the equations below :
What would be the mass of `Al_(2)(SO_(4))_(3)` formed per 27 g Al consumed?

Explanation

To find the mass of \( Al_2(SO_4)_3 \) formed per 27 g of Al consumed, we will follow these steps:

### Step 1: Write down the balanced chemical equation
The balanced chemical equation is:
\[ 2Al(s) + 3H_2SO_4(aq) \rightarrow Al_2(SO_4)_3(aq) + 3H_2(g) \]

### Step 2: Determine the molar mass of Aluminum (Al)
The atomic mass of Aluminum (Al) is 27 g/mol.

### Step 3: Calculate the number of moles of Aluminum in 27 g
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
For 27 g of Al:
\[ \text{Number of moles of Al} = \frac{27 \text{ g}}{27 \text{ g/mol}} = 1 \text{ mole} \]

### Step 4: Use the stoichiometry from the balanced equation
From the balanced equation, 2 moles of Al produce 1 mole of \( Al_2(SO_4)_3 \). Therefore, 1 mole of Al will produce:
\[ \frac{1 \text{ mole of } Al}{2 \text{ moles of } Al} \times 1 \text{ mole of } Al_2(SO_4)_3 = \frac{1}{2} \text{ moles of } Al_2(SO_4)_3 \]

### Step 5: Calculate the molar mass of \( Al_2(SO_4)_3 \)
The molar mass of \( Al_2(SO_4)_3 \) is calculated as follows:
- Aluminum (Al): \( 27 \text{ g/mol} \times 2 = 54 \text{ g/mol} \)
- Sulfur (S): \( 32 \text{ g/mol} \times 3 = 96 \text{ g/mol} \)
- Oxygen (O): \( 16 \text{ g/mol} \times 12 = 192 \text{ g/mol} \) (since there are 4 O atoms in each SO4 and 3 SO4 groups)

Adding these together:
\[ 54 + 96 + 192 = 342 \text{ g/mol} \]

### Step 6: Calculate the mass of \( Al_2(SO_4)_3 \) produced from 1 mole of Al
Since 1 mole of Al produces \( \frac{1}{2} \) moles of \( Al_2(SO_4)_3 \):
\[ \text{Mass of } Al_2(SO_4)_3 = \frac{1}{2} \text{ moles} \times 342 \text{ g/mol} = 171 \text{ g} \]

### Conclusion
The mass of \( Al_2(SO_4)_3 \) formed per 27 g of Al consumed is **171 g**.

---