Question

1. \\(\boldsymbol{4}\\) al + \\(\boldsymbol{3}\\) \\(\ce{o_2}\\) \\(\

ightarrow\\) \\(\boldsymbol{2}\\) \\(\ce{al_2o_3}\\) if 50.0 g of \\(\ce{al_2o_3}\\) is produced, what mass of al reacted?

Explanation:

Step1: Find molar mass of \( \text{Al}_2\text{O}_3 \) and \( \text{Al} \)

Molar mass of \( \text{Al} = 26.98 \, \text{g/mol} \), molar mass of \( \text{O} = 16.00 \, \text{g/mol} \).
Molar mass of \( \text{Al}_2\text{O}_3 = 2\times26.98 + 3\times16.00 = 101.96 \, \text{g/mol} \).

Step2: Calculate moles of \( \text{Al}_2\text{O}_3 \)

Moles of \( \text{Al}_2\text{O}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{50.0 \, \text{g}}{101.96 \, \text{g/mol}} \approx 0.4903 \, \text{mol} \).

Step3: Relate moles of \( \text{Al}_2\text{O}_3 \) to \( \text{Al} \) (from balanced equation \( 4\text{Al} + 3\text{O}_2

ightarrow 2\text{Al}_2\text{O}_3 \))
Mole ratio of \( \text{Al} : \text{Al}_2\text{O}_3 = 4 : 2 = 2 : 1 \).
Moles of \( \text{Al} = 2 \times \text{moles of } \text{Al}_2\text{O}_3 = 2 \times 0.4903 \, \text{mol} = 0.9806 \, \text{mol} \).

Step4: Calculate mass of \( \text{Al} \)

Mass of \( \text{Al} = \text{moles} \times \text{molar mass} = 0.9806 \, \text{mol} \times 26.98 \, \text{g/mol} \approx 26.4 \, \text{g} \).

Answer:

The mass of \( \text{Al} \) reacted is approximately \( \boldsymbol{26.4 \, \text{g}} \).