# What is the balanced equation for combustions of propane gas that yields carbon dioxide and water?

May 15, 2014

A combustion reaction is a reaction between a hydrocarbon burned in oxygen to form carbon dioxide and water.

A hydrocarbon is a molecule composed of carbon and hydrogen in this case propane. The pro- prefix tells us we have 3 carbon.
For hydrocarbons that end in the suffix -ane the formula is ${C}_{n} {H}_{2 n + 2}$. This means we need 8 hydrogen, making the formula ${C}_{3} {H}_{8}$.

Now we can format the basic reaction.

${C}_{3} {H}_{8} + {O}_{2} \to C {O}_{2} + {H}_{2} O$

A reminder that oxygen is a diatomic molecule as a gas, ${O}_{2}$.

Begin by balancing the hydrogen by adding a coefficient of 4 in from of the water.

${C}_{3} {H}_{8} + {O}_{2} \to C {O}_{2} + 4 {H}_{2} O$

This balances the hydrogen at 8. Now add a a coefficient of 3 in from of the carbon dioxide.

${C}_{3} {H}_{8} + {O}_{2} \to 3 C {O}_{2} + 4 {H}_{2} O$

This balances the carbon at 3. We now have 10 oxygen on the on the product side, 6 from $C {O}_{2}$ and 4 from ${H}_{2} O$. We need to add a coefficient of 5 in front of the oxygen to balance the oxygen at 10.

${C}_{3} {H}_{8} + 5 {O}_{2} \to 3 C {O}_{2} + 4 {H}_{2} O$

The combustion reaction is now balanced.