Enthalpy - Chemistry

Question

The combustion of propane is given by the following reaction.

C₃H₈ + 3 O₂ → 3 CO₂ + 4 H₂O

The enthalpy of reaction is −2202.0 ^kJ/_mol. How much energy (in joules) will be released if 43.11 grams of propane is burned. (Molar mass of propane = 44.11 g/mol). 

Answer 1

C₃H₈ + 3 O₂ → 3 CO₂ + 4 H₂O ... balanced equation

First find out how many moles of propane you have:

molar mass propane (C₃H₈) = 44.11 g / mol

43.11 g x 1 mol / 44.11 g = 0.9773 mols

Next, use 2202.0 kJ/mol to find the amount of kJ

2202.0 kJ / mol x 0.9773 mol = 2152 kJ