For the following reaction, you obtained 7.2 g of sodium sulfate, starting from 10 g of sulfuric acid. Sodium hydroxide is in excess. What is the % yield? H2SO4 + 2NaOH → Na2SO4 + 2H2O
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Identify the limiting reagent: you have been told sodium hydroxide is in excess, so you know sulfuric acid is the limiting reagent
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Calculate the moles of the limiting reagent: n = m/M. m = 10 g, M = (1 x 2) + 32.1 + (16 x 4) = 98.1 g mol-1. n = 10/98.1 = 0.10 mol
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The ratio of sulfuric acid to sodium sulfate is 1:1, so expected yield of sodium sulfate is 0.10 mol
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To work out your actual yield, calculate moles of sodium sulfate. n = m/M. m = 7.2 g, M = (23.0 x 2) + 32.1 + (16.0 x 4) = 142.1 g mol-1. n = 7.2/142.1 = 0.05 mol
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% yield = (actual/expected) x 100 = 50%
RB
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