What is the approximate percent yield for this reaction?

Ba (NO3) 2 + Na2SO4 - > BaSO4 + 2 NaNO3

The percent yield for this reaction is 91%.

Explanation:

Ba (NO3) 2 + Na2SO4 - > BaSO4 + 2NaNO3

Assume that 0.45 mol of Ba (NO3) 2 reacts with excess Na2SO4 to create 0.41 mol of BaSO4. Convert to grams of BaSO4 utilizing its molar mass (233.38 g/mol), which will be giving the actual yield. Actual Yield = 0.41 mol BaSO4 x (233.38 g BaSO4/1 mol BaSO4)

= 96 g BaSO4

Beginning with 0.45 mol of Ba (NO3) 2, compute the theoretical yield of BaSO4. Mole proportion 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%.