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29 May, 16:17

If you obtained 2.50 moles of CuSO4 from the reaction of CuO and H2SO4, but the theoretical yield was 3.19 moles, what was the percent yield?

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  1. 29 May, 19:10
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    78.4%

    Explanation:

    Just as context, write the chemical equation and the mole ratios

    1) Balanced chemical equation:

    CuO (s) + H₂SO₄ (aq) → CuSO₄ (aq) + H₂O (l)

    2) Therotetical (stoichiometric) mole ratios:

    1 mol CuO : 1 mol H₂SO₄ : 1 mol CuSO₄ : 1 mol H₂O

    You can calculate the percent yield from the amount of CuSO₄ obtained and the theoretical yield

    3) Percent yield

    Definition:

    Percent yield = (actual yield / theoretical yield) * 100

    Theoretical yiedl (given) : 3.19 moles CuSO₄

    Actual yield (given) : 2.50 moles CuSO₄

    Calcualtion:

    Substitute the values in the formula:

    Percent yield = (2.50 moles CuSO₄ / 3.19 moles CuSO₄) * 100 = 78.4%
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