Question

4) If 300.0 g of butane were completely combusted in the presence of excess oxygen, what mass of CO2 would be produced?

Answer 1

Molar mass of C4H10,

MM = 4*MM(C) + 10*MM(H)

= 4*12.01 + 10*1.008

= 58.12 g/mol

mass(C4H10)= 300.0 g

use:

number of mol of C4H10,

n = mass of C4H10/molar mass of C4H10

=(3*10^2 g)/(58.12 g/mol)

= 5.162 mol

Balanced chemical equation is:

2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O

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 = (8/2)* moles of C4H10

= (8/2)*5.1617

= 20.6469 mol

use:

mass of CO2 = number of mol * molar mass

= 20.65*44.01

= 9.087*10^2 g

Answer: 909 g