Question

Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as follows: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)Cu2S(s) + O2(g)  →  2Cu(s) + SO2(g)

What is the percent yield for the reaction if 45 grams of copper(I) sulfide reacts with excess What is the percent yield for the reaction if 45 grams of copper(I) sulfide reacts with excess

oxygen to produce an actual yield of 30 grams of copper?oxygen to produce an actual yield of   30   grams of copper?

Select one:

a. 84%

b. 77%

c. 69%

d. 59%

Answer 1

Atomic weights : Cu= 63.45, O= 16 and S=32

Molar masses : Cu2S= 2*63.45+32= 158.9, O2= 32, SO2= 32+2*16= 64

The reaction is Cu2S(s)+O2 (g)-----à2Cu(s)+SO2(g)

The reaction suggests 1 mole of Cu2S reacts with 1 mole of oxygen to produce 2 moles of Cu and 1 mole of SO2.

Since oxygen is excess, Cu2S will be the limiting reactants and moles of Cu formed will be decided by moles of Cu formed.

The reaction also suggests

158.9 gm of Cu2S reacts with excess oxygen to produce 2*63.45= 126.9 gm of CU

45 gm of Cu reacts with excess oxygen to produce 45*126.9/158.9 gm of Cu=35.93 gm of Cu

Hence theoretical yield of Cu= 35.94 gm

Actual yield = 30 gm

% yield= 100* actual yield/theoretical yield= 100*30/35.94=83.47% ( close answer is 84% , a