Question

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.

Answer 1

5.242

Step-by-step 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