I NEED HELP PLEASE

Question

Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 27. g of octane is mixed with 46.1 g of oxygen. Calculate the minimum mass of octane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Octane = C₈H₁₈

2C₈H₁₈ + 25O₂ ==> 16CO₂ + 18H₂O .. balanced equation

molar mass octane = 114 g / mole

moles octane = 27. g x 1 mol / 114 g = 0.237 moles

molar mass O₂ = 32 g / mol

moles O₂ = 46.1 g x 1 mol / 32 g = 1.44 moles

Limiting reactant is O₂ since it takes 12.5 x as much O₂ as octane, but we don't have that amount.

moles of octane consumed = 1.44 moles O₂ x 2 moles octane / 25 moles O₂ = 0.115 moles octane used up

moles octane remaining = 0.237 moles - 0.115 moles = 0.122 moles octane remaining

mass octane remaining = 0.122 moles octane x 114 g / mole = 13.9 g = 14. g octane remaining (2 sig.figs.)