Question

How many grams of CO₂ are produced by the combustion of 344 g of a mixture that is 33.6% CH₄ and 66.4% C₃H₈ by mass?

Answer 1

Step 1: calculate CO2 produced from CH4.
Mass of CH4 = 33.6 % of 344 g
= 33.6 * 344 / 100
= 115.6 g

CH4 + 2 O2 ---> CO2 + 2 H2O

Molar mass of CH4,
MM = 1*MM(C) + 4*MM(H)
= 1*12.01 + 4*1.008
= 16.042 g/mol

mass of CH4 = 1.156*10^2 g
mol of CH4 = (mass)/(molar mass)
= 1.156*10^2/16.04
= 7.206 mol

According to balanced equation
mol of CO2 formed = moles of CH4
= 7.206 mol


Molar mass of CO2,
MM = 1*MM(C) + 2*MM(O)
= 1*12.01 + 2*16.0
= 44.01 g/mol


mass of CO2 = number of mol * molar mass
= 7.206*44.01
= 317.1 g

Step 2: calculate CO2 produced from C3H8.
Mass of C3H8 = total mass - mass of CH4
= 344 g - 115.6 g
= 228.4 g

C3H8 + 5 O2 ---> 3 CO2 + 4 H2O

Molar mass of C3H8,
MM = 3*MM(C) + 8*MM(H)
= 3*12.01 + 8*1.008
= 44.094 g/mol

mass of C3H8 = 2.284*10^2 g
mol of C3H8 = (mass)/(molar mass)
= 2.284*10^2/44.09
= 5.18 mol



According to balanced equation
mol of CO2 formed = (3/1)* moles of C3H8
= (3/1)*5.18
= 15.54 mol


Molar mass of CO2,
MM = 1*MM(C) + 2*MM(O)
= 1*12.01 + 2*16.0
= 44.01 g/mol


mass of CO2 = number of mol * molar mass
= 15.54*44.01
= 683.9 g

Step 3:
Total mass of CO2 produced = 317.1 g + 683.9 g
= 1001 g
Answer: 1000. g