Question

The reaction of 9.15 g of carbon with excess O2 yields 8.25 g of CO2. What is the percent yield of this reaction?

Answer 1

Molar mass of C = 12.01 g/mol

mass(C)= 9.15 g

use:

number of mol of C,

n = mass of C/molar mass of C

=(9.15 g)/(12.01 g/mol)

= 0.7619 mol

Balanced chemical equation is:

C + O2 ---> CO2

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

According to balanced equation

mol of CO2 formed = (1/1)* moles of C

= (1/1)*0.7619

= 0.7619 mol

use:

mass of CO2 = number of mol * molar mass

= 0.7619*44.01

= 33.53 g

% yield = actual mass*100/theoretical mass

= 8.25*100/33.53

= 24.6 %

Answer: 24.6 %