Liquid octane (CH3(CH2)6 CH3) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O).

Question

Suppose 10. g of octane is mixed with 61.2 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 2 significant digits.

= ______ g

Answer 1

Octane can be written as C8H18.

The balanced equation for combustion of octane is:

2C₈H₁₈ + 25 O₂ ==> 16CO₂ + 18H₂O

Finding the limiting reactant:

10.g C₈H₁₈ x 1 mol C₈H₁₈ / 118 g = 0.0847 moles (÷2->0.042)

61.2 g O₂ x 1 mol O₂ / 32 g = 1.91 moles (÷25->0.077)

C₈H₁₈ is the limiting reactant and will dictate the maximum amount of H₂O that can form.

Maximum H2O formed:

0.0847 moles C₈H₁₈ x 18 moles H₂O / 2 moles C₈H₁₈ x 18 g H₂O/mol = 14 g H₂O formed (2 sig.figs.)