The percent yield for this reaction is 91%.
Explanation:
Ba(NO3)2 + Na2SO4 — > BaSO4 + 2NaNO3
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Assume that 0.45 mol of Ba(NO3)2 reacts with excess Na2SO4 to create 0.41 mol of BaSO4.
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Convert to grams of BaSO4 utilizing its molar mass (233.38 g/mol), which will be giving the actual yield.
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Actual Yield = 0.41 mol BaSO4 x (233.38 g BaSO4/1 mol BaSO4)
= 96 g BaSO4
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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.
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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%.
