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%.

The answer is the first option just took the test