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.

The most plausible answer is b, duration. the given amount of time is considered how long it last for.