Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction.

Question

Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 8.0 g of octane is mixed with 8.96 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

This is not much different from your previous questions. Here goes one more time.

BTW, octane should be CH₃(CH₂)₆CH₃ which is also C₈H₁₈

Write the correctly balanced equation for the reaction:

2C₈H₁₈ + 25 O₂ ==> 16CO₂ + 18H₂O ... balanced equation for combustion of octane

Since we are given the amounts of BOTH reactants, we must find which one is limiting.

moles octane = 8.0 g C₈H₁₈ x 1 mol/114 g = 0.070 moles C₈H₁₈

moles O₂ = 8.96 g O₂ x 1 mol/32 g = 0.28 moles O₂ --> LIMITING REACTANT

max mass CO₂ = 0.28 mol O₂ x 16 mol CO₂ / 25 mol O₂ x 44 g CO₂ / mol CO₂ = 7.9 g CO₂ (2 sig. figs.)