How many grams of Solid Aluminium Sulfide can be prepared by the reaction of 10.0 g of Aluminium and 15.0 g of Sulfur. How many grams of the non-limiting reactant are in excess.

Aluminum reacts with sulfur to form aluminum sulfide. Aluminum typically forms a $\text{Al}^{3+}$ ion and sulfur forms a $\text{S}^{2-}$ ion, so the compound is $\text{Al}_2\text{S}_3$.

• Molar mass of Al = $26.98 \text{ g/mol}$
• Molar mass of S = $32.07 \text{ g/mol}$
• Molar mass of $\text{Al}_2\text{S}_3 = (2 \times 26.98 \text{ g/mol}) + (3 \times 32.07 \text{ g/mol}) = 53.96 \text{ g/mol} + 96.21 \text{ g/mol} = 150.17 \text{ g/mol}$

• Moles of Al = $\frac{10.0 \text{ g}}{26.98 \text{ g/mol}} = 0.3706 \text{ mol}$
• Moles of S = $\frac{15.0 \text{ g}}{32.07 \text{ g/mol}} = 0.4677 \text{ mol}$

To find the limiting reactant, divide the moles of each reactant by its stoichiometric coefficient from the balanced equation:

• For Al: $\frac{0.3706 \text{ mol}}{2} = 0.1853$
• For S: $\frac{0.4677 \text{ mol}}{3} = 0.1559$ Since 0.1559 is less than 0.1853, Sulfur (S) is the limiting reactant, and Aluminum (Al) is the excess reactant.

The amount of product is determined by the limiting reactant (Sulfur). From the balanced equation, 3 moles of S produce 1 mole of $\text{Al}_2\text{S}_3$.

• Moles of $\text{Al}_2\text{S}_3$ produced = $0.4677 \text{ mol S} \times \frac{1 \text{ mol Al}_2\text{S}_3}{3 \text{ mol S}} = 0.1559 \text{ mol Al}_2\text{S}_3$
• Mass of $\text{Al}_2\text{S}_3$ = $0.1559 \text{ mol} \times 150.17 \text{ g/mol} = 23.41 \text{ g}$ Rounding to three significant figures, the mass of $\text{Al}_2\text{S}_3$ produced is $\boxed{\text{23.4 g}}$.

First, calculate the moles of Al consumed by the limiting reactant (S). From the balanced equation, 3 moles of S react with 2 moles of Al.

• Moles of Al consumed = $0.4677 \text{ mol S} \times \frac{2 \text{ mol Al}}{3 \text{ mol S}} = 0.3118 \text{ mol Al}$
• Mass of Al consumed = $0.3118 \text{ mol} \times 26.98 \text{ g/mol} = 8.413 \text{ g Al}$
• Mass of Al in excess = Initial mass of Al - Mass of Al consumed
• Mass of Al in excess = $10.0 \text{ g} - 8.413 \text{ g} = 1.587 \text{ g}$ Rounding to three significant figures, the mass of Aluminum in excess is $\boxed{\text{1.59 g}}$.