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March 25, 2017

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13. Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as
follows:
Cu2S(s) + O2(g) „_ 2Cu(s) + SO2(g)
If the reaction of 0.540 kg of copper(I) sulfide with excess oxygen produces 0.140 kg of
copper metal, what is the percent yield?
A) 32.5%
B) 25.9%
C) 64.9%
D) 130%
E) 39.9%

  • chemistry 21 - ,

    If that reaction were to go to completion, one mole of Cu2S(weighing 159 g) would form 2 moles of Cu (weighing 127g). 0.540 kg of Cu2S would thus lead to production of
    0.540 * (127/159) = 0.431 g of Cu.

    The 0.140 g actually produced is 32.5% of that amount. That is the percent yield.

  • chemistry 21 - ,

    Convert 0.540 kg Cu2S to moles. Moles = grams/molar mass.

    Using the coefficients in the balanced equation, convert moles Cu2S to moles Cu metal.

    Now convert moles Cu metal to grams. grams = moles x molar mass. This is the theoretical yield.

    % yield = (0.140 kg/theoretical yield)*100 = ??
    Be sure numerator and denominator are in the same units.

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