Question

A reaction vessel contains 13.1 g of CO and 13.1g of CO2. How many grams of CO2 could be produced according to the following unbalanced reaction? 

 CO+O2→CO2

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)= 13.1 g

use:

number of mol of CO,

n = mass of CO/molar mass of CO

=(13.1 g)/(28.01 g/mol)

= 0.4677 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 13.1 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(13.1 g)/(32 g/mol)

= 0.4094 mol

Balanced chemical equation is:

2 CO + O2 ---> 2 CO2 +

2 mol of CO reacts with 1 mol of O2

for 0.4677 mol of CO, 0.2338 mol of O2 is required

But we have 0.4094 mol of O2

so, CO is limiting reagent

we will use CO in further calculation

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)*0.4677

= 0.4677 mol

use:

mass of CO2 = number of mol * molar mass

= 0.4677*44.01

= 20.58 g

Answer: 20.6 g