the balanced equation for the above reaction is;
Ca + S ---> CaS
stoichiometry of Ca to S is 1:1
Number of Ca moles - 1.6 g/ 40 g/mol = 0.04 mol
Number of S moles - 4.25 g / 32 g/mol = 0.13 mol
since Ca to S molar ratio is 1:1, there's excess amount of S present.
Ca is the limiting reactant.
stoichiometry of Ca to CaS is 1:1
Number of CaS moles to be formed - 0.04 mol
Therefore mass of CaS - 0.04 mol x 72 g/mol = 2.88 g
therefore theoretical yield is 2.88