Consider the combustion of octane (C8H18)

2C8H18+25O2-> 16CO2+18H2O

How many grams of CO2 are produced when 191.6g of octane are burned?

The balanced equation for the above reaction is as follows;

2C₈H₁₈ + 25O₂ - - - > 16CO₂ + 18H₂O

stoichiometry of octane to CO₂ is 2:16

number of C₈H₁₈ moles reacted - 191.6 g / 114 g/mol = 1.68 mol

when 2 mol of octane reacts it forms 16 mol of CO₂

therefore when 1.68 mol of octane reacts - it forms 16/2 x 1.68 = 13.45 mol of CO₂

number of CO₂ moles formed - 13.45 mol

therefore mass of CO₂ formed - 13.45 mol x 44 g/mol = 591.8 g

mass of CO₂ formed is 591.8 g