Question

Calcium reacts with sulfur (S8) forming calcium sulfide. What is the theoretical yield (g) of CaS(s) that could be prepared from 1.83 g of Ca(s) and 3.35 g of sulfur(s)? Enter your answer in decimal format with two decimal places and no units.

Answer 1

First you need a balanced equation. In stoichiometry, you ALWAYS need a balanced equation.

Ca + S --> CaS

Easy enough. Now you need to know the molar masses of the involved substances:

Ca = 40.08 g/mol
S = 32.07 g/mol
CaS = 40.08 + 32.07 = 72.15 g/mol

Next, you're given the starting masses of two reactants. Since we don't know if those masses are stoichiometrically equal, we're going to have to determine which is the limiting reactant. We can do that by converting both to moles:

1.83 g Ca x (1 mol Ca)/(40.08 g Ca) = 0.0457 mol Ca

3.35 g S x (1 mol S)/(32.07 g S) = 0.1045 mol S

Since there are more moles of sulfur than calcium (and since calcium and sulfur react in a 1:1 ratio in this reaction), you know that the calcium will be used up first. Therefore, calcium is the limiting reactant.

Now you're ready to predict the theoretical yield of CaS that can be formed. You know that you have 0.0457 moles of Ca available, and the ratio of CaS formed to Ca used is 1:1. Therefore, you'll also get 0.0457 mol CaS formed. Now just convert from moles of calcium sulfide to grams.

0.0457 mol CaS x (72.15 g CaS)/(1 mol CaS) = 3.2972 ≈ 3.30 g of CaS