Chemistry Forum

Calculate the mass of carbon dioxide produced from 22.6 g of octane, C₈H₁₈, in the following reaction. 2C₈H₁₈(g) + 25O₂(g) --> 16CO₂(g) + 18H₂O (g)

69.8 grams

you have 22.6 / (8*12 + 18 ) = 0.198 moles of C₈H₁₈ and from the equation the amount of moles of CO₂ is 8times higher so in the end you have
8 * 0.198 moles of CO₂ which is 1.58 mole.
1.58 * (12+32) = 69.8g

In general:


Where
Mp : Mass of Product (CO2)
Wp : Atomic-Weight of Product = 44.009 g/mole
Np : Stocimetric Number of Product (ie. moles in the equation) = 16
Mr : Mass of Reactant (C8H18) = 22.6 g
Wr : Atomic-Weight of Reactant = 114.23 g/mole
Nr : Stocimetric Number of Reactant = 2

Mp = 69.656 g