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

1,002.936 g

Explanation:

The combustion equation of each will be;

CH4 + 2O2 = CO2 + H2O

C3H8 + 5O2 = 3CO2 + 4H2O

We are told the mixture is 344g.

Thus;

For CH4 combustion;

Amount of CH4 = 0.336 × 344 = 115.584g

Molar mass of CH4 is 16 g/mol

Number of moles of CH4 is;

n = 115.584g/(16 g/mol)

n = 7.224 moles

n(CO2) = n(CH4) = 7.224 moles

For C3H8 combustion;

Amount of C3H8 = 0.664 × 344 = 228.416 g

Molar mass of C3H8 = 44 g/mol

Thus;

Number of moles of C3H8 = 228.416 g/(44 g/mol) = 5.19 mol

n(CO2) = 3n(C3H8) = 3 × 5.19 = 15.57 moles

Total moles of CO2 = 7.224 moles + 15.57 moles = 22.794 mol

Molar mass of CO2 = 44 g/mol

Thus amount of CO2 by mass = 22.794 mol × 44 g/mol = 1,002.936 g