chemistry problem

Question

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

0.41 mol of BaSO4 was produced. Convert to grams of BaSO4 using its molar mass (233.38 g/mol), which will be the actual yield.

Actual Yield = 0.41 mol BaSO4 x (233.38 g BaSO4 / 1 mol BaSO4) = 96 g BaSO4

Starting with 0.45 mol of Ba(NO3)2, calculate the theoretical yield of BaSO4. Mole ratio of Ba(NO3)2 to BaSO4 is 1:1.

0.45 mol Ba(NO3)2 x (1 mol BaSO4 / 1 mol Ba(NO3)2) x (233.38 g BaSO4 / 1 mol BaSO4) = 105 g BaSO4

%-Yield = (Actual Yield / Theoretical Yield) x 100%

= (96 / 105) x 100% = 91%