Question

32. Refer to the following unbalanced equation: C6H14 + O2 ® CO2 + H2O
What mass of carbon dioxide (CO2) can be produced from 11.2 g of C6H14 and excess oxygen?

Answer 1

Molar mass of C6H14,

MM = 6*MM(C) + 14*MM(H)

= 6*12.01 + 14*1.008

= 86.172 g/mol

mass(C6H14)= 11.2 g

use:

number of mol of C6H14,

n = mass of C6H14/molar mass of C6H14

=(11.2 g)/(86.17 g/mol)

= 0.13 mol

Balanced chemical equation is:

2 C6H14 + 19 O2 ---> 12 CO2 + 14 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 = (12/2)* moles of C6H14

= (12/2)*0.13

= 0.7798 mol

use:

mass of CO2 = number of mol * molar mass

= 0.7798*44.01

= 34.32 g

Answer: 34.3 g