Question

1. if 77.32 grams of s are used, how many moles of al are used?

Explanation:

To solve this problem, we first need the balanced chemical equation for the reaction involving aluminum (Al) and sulfur (S). The reaction is typically the formation of aluminum sulfide:

$$2\text{Al} + 3\text{S} ightarrow \text{Al}_2\text{S}_3$$

Step 1: Calculate moles of S

The molar mass of S is approximately \( 32.07 \, \text{g/mol} \). Use the formula:

$$\text{Moles of S} = \frac{\text{Mass of S}}{\text{Molar mass of S}}$$

Substitute \( \text{Mass of S} = 77.32 \, \text{g} \) and \( \text{Molar mass of S} = 32.07 \, \text{g/mol} \):

$$\text{Moles of S} = \frac{77.32 \, \text{g}}{32.07 \, \text{g/mol}} \approx 2.411 \, \text{mol}$$

Step 2: Relate moles of S to moles of Al (using stoichiometry)

From the balanced equation, the mole ratio of \( \text{Al} : \text{S} \) is \( 2 : 3 \). This means:

$$\text{Moles of Al} = \text{Moles of S} \times \frac{2 \, \text{mol Al}}{3 \, \text{mol S}}$$

Substitute \( \text{Moles of S} \approx 2.411 \, \text{mol} \):

$$\text{Moles of Al} = 2.411 \, \text{mol S} \times \frac{2}{3} \approx 1.607 \, \text{mol}$$

Answer:

Approximately \( \boldsymbol{1.61 \, \text{moles}} \) of Al are used (or \( 1.607 \, \text{mol} \) for more precision).