Question

2 al + 3 s → al₂s₃

1. if 3.12 moles of al₂s₃ are used, how many grams of s are also used?

Explanation:

Step1: Determine mole ratio of S to \( \text{Al}_2\text{S}_3 \)

From the balanced equation \( 2\text{Al} + 3\text{S}
ightarrow \text{Al}_2\text{S}_3 \), the mole ratio of \( \text{S} \) to \( \text{Al}_2\text{S}_3 \) is \( \frac{3\ \text{mol S}}{1\ \text{mol}\ \text{Al}_2\text{S}_3} \).

Step2: Calculate moles of S

Given moles of \( \text{Al}_2\text{S}_3 = 3.12\ \text{mol} \).
Moles of S \( = 3.12\ \text{mol}\ \text{Al}_2\text{S}_3 \times \frac{3\ \text{mol S}}{1\ \text{mol}\ \text{Al}_2\text{S}_3} = 9.36\ \text{mol S} \).

Step3: Convert moles of S to grams

Molar mass of S is \( 32.07\ \text{g/mol} \).
Mass of S \( = 9.36\ \text{mol} \times 32.07\ \text{g/mol} \approx 299.97\ \text{g} \) (or ~300 g).

Answer:

Approximately \( 300\ \text{g} \) (or more precisely \( \approx 299.97\ \text{g} \)) of S are used.