Consider the combustion reaction of propane: C₃H₈(g)+ 5 O₂(g)→ 3 CO₂(g)+ 4 H₂O(g),where ΔH = -531 kcal. If 6.70 × 10⁴ kcal of energy is released in the reaction,how many grams of oxygen were consumed?

Question

Quizplus · Accepted Answer

The reaction releases 531 kcal per mole of propane. To release 6.70 × 10^4 kcal, (6.70 × 10^4 kcal) / (531 kcal/mol) = 126.17 moles of propane are needed. Since 5 moles of O2 are required per mole of propane, 126.17 moles of propane require 630.85 moles of O2. The molar mass of O2 is 32 g/mol, so 630.85 moles × 32 g/mol = 2.02 × 10^4 g of O2.

Consider the Combustion Reaction of Propane: C3H8(g)+ 5 O2(g)→ 3

Consider the Combustion Reaction of Propane: C₃H₈(g)+ 5 O₂(g)→ 3