Question

Hydrocarbon mixtures are used as fuels. How many grams of CO₂(g) are produced by the combustion of 574 g of a mixture that is 31.6% CH₄ and 68.4%  C₃H₈ by mass?

___ × 10____ g

(Enter your answer in scientific notation.)

Answer 1

balance equation

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

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

Work out mass then moles of each hydrocarbon

mass CH4 = 31.6/100 x 574 g
= 181.384 g

moles = mass / molar mass
moles CH4 =181.384 g / 16.042 g/mol
moles CH4 = 11.30 moles

mass C3H8 = 68.4 / 100 x 574 g
= 392.616 g

moles C3H8 = 392.616 g / 44.094 g/mol
moles C3H8 = 8.90 moles

how much CO2 forms from each

1 mol CH4 produces 1 moles CO2
So 11.30 mol CH4 produces 11.30 mol CO2

1 mole C3H8 produces 3 moles CO2
Thus 8.90 moles C3H8 gives (3 x 8.90) mol CO2
= 26.7122 mol CO2

Total moles CO2 = 111.30 + 26.72 mol = 38.01 mol CO2

mass = molar mass x moles
mass CO2 = 44.01 g/mol x 38.01 mol
mass CO2 = 1672.53 g
= 1.67*10^3 gram