45.2 grams of aluminum nitrate are reacted with 71.9 mL of a 0.880 M solution of calcium sulfide. The balanced chemical equation can be seen below:

LaTeX: 2Al\left(NO_3\right)_3\:+\:3\:CaS\: \longrightarrow\:Al_2S_3\:+\:3Ca\le ft(NO_3\right)_22 A l ( N O 3 ) 3 + 3 C a S ⟶ A l 2 S 3 + 3 C a ( N O 3 ) 2

If 2.95 grams of precipitate is filtered out of the resulting solution, what is the actual yield?