Astudent mixes a solution containing 10.0 g bacl2 (m = 208.2) with a solution containing 10.0 g na2so4 (m = 142.1) and obtains 12.0 g baso4 (m = 233.2). what is the percent yield of this reaction?

The balanced equation for the above reaction is as follows;
Na₂SO₄ + BaCl₂ --> BaSO₄ + 2NaCl
Na₂SO₄ reacts with BaCl₂ in the molar ratio 1:1
Number of Na₂SO₄ moles - 10.0 g / 142.1 g/mol = 0.0704 mol
Number of BaCl₂ moles - 10.0 g / 208.2 g/mol = 0.0480 mol
this means that 0.0480 mol of each reactant is used up, BaCl₂ is the limiting reactant and Na₂SO₄ has been provided in excess. 
stoichiometry of BaCl₂ to BaSO₄ is 1:1
number of BaSO₄ moles formed - 0.0480 mol
Mass of BaSO₄ - 0.0480 mol x 233.2 g/mol = 11.2 g
theoretical yield is 11.2 g but the actual yield is 12.0 g
the actual product maybe more than the theoretical yield of the product as the measured mass of the actual yield might contain impurities. 
percent yield - 12.0 g/ 11.2 g x 100% = 107% 
this is due to impurities present in the product or product could be wet.

Sorry i dont understand the question can you rephase it?

H tricky one let me think and get back to u