Calcium reacts with sulfur forming calcium sulfide. What is the theoretical yield (g) of CaS (s) that could be prepared from 8.54 g of Ca (s) and 2.33 g of sulfur (s) ? Do not type units with your answer.

5.242

Explanation:

Molar mass of Ca = 40.078 g/mol

Molar mass of S = 32.065 g/mol

Equation of reaction

Ca (s) + S (s) - CaS (s)

To find the limiting reagent

Mole of Ca = 8.54/40.078 (g/mol) = 0.213 mol

Mole of S = 2.33g / 32.065 (g/mol) = 0.072664 mol

From their mole ratio, sulphur is the limiting reagent

From the reaction

1 mole of calcium react with 1 mole of Sulphur to yield 1 mole of calcium sulphide

0.072664 mol of calcium will react with 0.072664 of sulphur to yield 0.072664 mole of CaS

Theoretical yield of CaS = 0.072664 * molar mass of CaS = 0.072664 * 72.143 g/mol = 5.242 g