Question

Pure copper Amy be produced by the reaction of Copper (I) Sulfide with oxygen as follows

Cu2S(s) + O2(g)  →  2Cu(s) + SO2(g)

If the reaction of .530kg copper(I) sulfide with excess oxygen produces 0.290 kg of copper metal, what is the percent yield?

Answer 1

Molar mass of Cu2S = 2*MM(Cu) + 1*MM(S)

= 2*63.55 + 1*32.07

= 159.17 g/mol

mass of Cu2S = 530.0 g

we have below equation to be used:

number of mol of Cu2S,

n = mass of Cu2S/molar mass of Cu2S

=(530.0 g)/(159.17 g/mol)

= 3.33 mol

we have the Balanced chemical equation as:

Cu2S + O2 ---> 2 Cu + SO2

Molar mass of Cu = 63.55 g/mol

From balanced chemical reaction, we see that

when 1 mol of Cu2S reacts, 2 mol of Cu is formed

mol of Cu formed = (2/1)* moles of Cu2S

= (2/1)*3.3298

= 6.6595 mol

we have below equation to be used:

mass of Cu = number of mol * molar mass

= 6.66*63.55

= 4.232*10^2 g

% yield = actual mass*100/theoretical mass

= 290*100/423.2142

= 68.5 %

Answer: 68.5 %