A student 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.