Question

2 al + 3 s → al₂s₃ (13) if 3.12 moles of al₂s₃ are used, how many grams of s are also used?

Explanation:

Step1: Determine the mole ratio of S to \( Al_2S_3 \)

From the balanced chemical equation \( 2Al + 3S
ightarrow Al_2S_3 \), the mole ratio of \( S \) to \( Al_2S_3 \) is \( \frac{3\ mol\ S}{1\ mol\ Al_2S_3} \).

Step2: Calculate moles of S

Given moles of \( Al_2S_3 = 3.12\ mol \). Using the mole ratio, moles of \( S = 3.12\ mol\ Al_2S_3 \times \frac{3\ mol\ S}{1\ mol\ Al_2S_3} = 9.36\ mol\ S \).

Step3: Calculate mass of S

Molar mass of \( S = 32.07\ g/mol \). Mass of \( S = moles \times molar\ mass = 9.36\ mol \times 32.07\ g/mol \approx 299.97\ g \) (or approximately 300 g).

Answer:

Approximately 300 grams (or 299.97 grams) of S are used.