Question

The reaction of 9.05 g of carbon with excess O2 yields 110 g of CO2 What is the percent yield of this reaction?

Answer 1

Molar mass of C = 12.01 g/mol

mass of C = 9.05 g
mol of C = (mass)/(molar mass)
= 9.05/12.01
= 0.7535 mol


Balanced chemical equation is:
C + O2 —> CO2


According to balanced equation
mol of CO2 formed = moles of C
= 0.7535 mol


Molar mass of CO2,
MM = 1*MM(C) + 2*MM(O)
= 1*12.01 + 2*16.0
= 44.01 g/mol


mass of CO2 = number of mol * molar mass
= 0.7535*44.01
= 33.16 g

% yield = actual mass*100/theoretical mass
= 11*100/33.16
= 33.2 %
Answer: 33.2 %