When 53.8 g calcium and 32.7 g nitrogen gas undergo a reaction that has a yield of 87.0%, what mass of calcium?

The limiting reactant (the one in shortest supply) is the one that will produce the smallest amount of product.

53.8 g Ca x (1 mole Ca / 40.1 g Ca) x (1 mole Ca3N2 / 3 moles Ca) x (148.3 g Ca3N2 / 1 mole Ca3N2 ) = 66.3 g Ca3N2

32.7 g N2 x (1 mole N2 / 28.0 g N2) x (1 mole Ca3N2 / 1 mole N2) x (148.3 g Ca3N2 / 1 mole Ca3N2) = 173 g Ca3N2

Obviously the Ca produces less product, so it is the limiting reactant. All of it gets used up and some N2 is left over. 66.3 g of Ca3N2 are produced theoretically.

Actual yield = theoretical yield x % yield =
66.3 g x 0.870 = 57.7 g Ca3N2

moles Ca = 53.8 g/ 40.078 g/mol= 1.34
moles N2 = 32.7 g/ 28.0134 g/mol=1.17

Ca is the limiting reactant ( 1,17 x3=3.51 moles are needed)
the ratio between Ca and Ca3N2 is 3 : 1
moles Ca3N2 = 1.34/3 =0.447
theoretical amount Ca3N2 = 0.447 mol x 148.25 g/mol=66.3 g

actual yield = 66.3 x 87.0/100=57.7 g