Posted by lee on Sunday, January 31, 2010 at 12:15am.
13. Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as
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?
chemistry 21 - drwls, Sunday, January 31, 2010 at 1:15am
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 - DrBo222, Sunday, January 31, 2010 at 1:18am
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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