If you start with 5.0 mL of acetone and 10 mL of benzaladehyde and obtain 5.07 grams of product, ...

Can you update us with the answer, I'm really struggling on this one too

acetone + benzaldehyde = dibenzalacetone + water
C3H6O + 2 C7H6O → C17H14O + 2 H2O

The second and third sources below give the densities of acetone and benzaldehyde as 0.791 g/cm^3 and
1.0415 g/ml, respectively.

(5.0 mL C3H6O) x (0.791 g/mL) / (58.0791 g C3H6O/mol) = 0.06810 mol C3H6O
(10 mL C7H6O) x (1.0415 g/ml) / (106.1219 g C7H6O/mol) = 0.09814 mol C7H6O

0.09814 mole of C7H6O would react completely with 0.09814 x (1/2) = 0.04907 mole of C3H6O, but there is more C3H6O present than that, so C3H6O is in excess and C7H6O is the limiting reactant.

(0.09814 mol C7H6O) x (1 mol C17H14O / 2 mol C7H6O) x (234.2925 g C17H14O/mol) =
11.5 g C17H14O in theory

(5.07 g) / (11.5 g) = 0.441 = 44.1% yield