Question

what is the percent yield of carbon dioxide if the reaction of 73.6 g of carbon monoxide produces 85.9 g of carbon dioxide?

Answer 1

Molar mass of CO,

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

= 1*12.01 + 1*16.0

= 28.01 g/mol

mass(CO)= 73.6 g

use:

number of mol of CO,

n = mass of CO/molar mass of CO

=(73.6 g)/(28.01 g/mol)

= 2.628 mol

Balanced chemical equation is:

2 CO + O2 ---> 2 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 = (2/2)* moles of CO

= (2/2)*2.6276

= 2.6276 mol

use:

mass of CO2 = number of mol * molar mass

= 2.628*44.01

= 115.6421 g

% yield = actual mass*100/theoretical mass

= 85.9*100/115.6421

= 74.2809 %

Answer: 74.3 %