What is the percent yield for this reaction?

Question

The equation below shows the reaction of barium nitrate (Ba(NO₃)₂) with sodium sulfate (Na₂SO₄). Assume that 0.45 mol of Ba(NO₃)₂ reacted with excess Na₂SO₄ to produce 0.41 mol of BaSO₄.

Ba(NO₃)₂ + Na₂SO₄ —> BaSO₄ + 2 NaNO₃

What is the approximate percent yield for this reaction?

Answer 1

Ba(NO₃)₂ + Na₂SO₄ ==> BaSO₄ + 2NaNO₃ ... balanced equation

Theoretical yield of BaSO4 from 0.45 moles of Ba(NO3):

0.45 moles Ba(NO3)2 x 1 mol BaSO4 / mol Ba(NO3)2 = 0.45 moles BaSO4

Percent yield = actual yield / theoretical yield (x100%) = 0.41 mol / 0.45 mol (x100%) = 91 %