Question #d02c8

Feb 6, 2017

${C}_{3} {H}_{8} + 5 {O}_{2} \to 3 C {O}_{2} + 4 {H}_{2} O$
Assuming complete combustion as in the balanced equation above, ie all carbon is oxidised to $C {O}_{2}$ and none to $C$ or $C O$, the 3 carbons will require 3 moles of oxygen (remember oxygen is ${O}_{2}$, not just $O$), and the 8 hydrogens will require four oxygens, i.e 2 moles of oxygen each. Adding those two up you get 5 moles of oxygen ie 5x${O}_{2}$). This is per mole of propane. If we have 4 moles of propane, we will need 4x5=20 moles of oxygen in total.
A quick shortcut to balance these types of combustion reactions (ie combustion of hydrocarbons) is to add the number of carbon atoms in the hydrocarbon to the number of hydrogens divided by four e.g pentane is ${C}_{5} {H}_{12}$, so it requires 5+(12/4)=5+3=8 moles of oxygen for complete combustion. The number of carbons equals the number of carbon dioxides formed and the number of hydrogens divided by two equals the number of waters.