How many grams of oxygen are required to react with 14grams of octane

Answer 1

This is a combustion reaction. The products of combustion are CO₂ + H₂O. You must first balance thje reaction: C₈H₁₈ + O₂ ---> CO₂ + H₂O. C₈H_{18 +} 25/2 O₂ ---> 8CO₂ + 9H₂O, multiply the entire equation by 2 to clear the 1/2 in the denominator of the O₂, and you get 2C₈H₁₈ + 25O₂---> 16CO₂ + 18H₂O. Now the equation is balanced so convert 14 grams of octane to moles, 14grams/114 grams/mole, and you have 0.8421 moles of octane. 0.8421 moles octane x 25 moles O₂/2 moles octane(from the balanced equation) =10.52 moles O₂=10.52 moles x 32 grams O₂/mole=336.84 grams O₂.