Ba(NO3)2+Na2SO4->BaSO4+2NaNO3 75g of Ba(NO3)2 were used and 63.45g BaSO4 were obtained. Calculate the theoretical yield and percent yield of BaSO4.
From the balanced equation above, we can see that 1 mole of barium nitrate produces one mole of barium sulfate. We know that 75 g of barium nitrate gave 63.45 g of barium sulfate. We should convert the barium nitrate reactant into moles:
75 g Ba(NO3)2 * (1 mole/261.3 g) = 0.287 moles Ba(NO3)2
Since the ratio of reactant to product is 1:1 as stated above, this means that 0.287 moles of BaSO4 is the theoretical amount of product produced. Let's convert that to grams:
0.287 moles BaSO4 * (233.4 g/1 mole) = 67 grams BaSO4
So the theoretical amount of BaSO4 produced if the reaction had gone 100% is 67 grams. Instead, we actually got 63.45 grams. So divide the two to get the percent yield:
63.45 g/67 g = 94.7 % yield.