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