what is the percent yield of the reaction between 5.2 mL of 0.1233 M barium nitrate and 4.9 mL of 0.1256 sodium sulfate if 0.140 grams of the precipitate are produced?

Formula?...
BaNO3(aq) + Na2SO4(aq) --> Na2NO3(aq) + BaSO4(s)

This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants.

mols Ba(NO3)2 = M x L = ?
mols Na2SO4 = M x L = ?

Using the coefficients in the balanced equation, convert mols Ba(NO3)2 to mols BaSO4.
Do the same for mols Na2SO4 to mols BaSO4.
It is likely that the two values will not agree which means one of them is not right; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR. Convert the smaller mols BaSO4 to grams by grams = mols BaSO4 x molar mass BaSO4. This is the theoretical yield (TY). The actual yield (AY) from the problem is 0.140g
%yield = (AY/TY)*100 = ?