One of the more common combustion reactions is the combustion of an alkane in the presence of oxygen.
A general strategy for balancing these combustion reactions is to first balance the carbon (C) atom, followed by the hydrogen atoms (H) and then then oxygen (O) atoms.
Remember: C H O
The combustion of alkanes in the presences of oxygen will produce carbon dioxide and water. Here is the reaction for the combustion of propane , an alkane containing 3 carbon atoms.
Here is the balanced equation for the combustion of propane , an alkane
containing 3 carbon atoms:
C3H8 + 5 O2 -----> 3 CO2 + 4 H2O
Let's examine how this combustion reaction was balanced.
Strategy for balancing a combustion reaction
1. Write the reaction.
C3H8 + O2 -----> CO2 + H2O
2. Next balance the carbon atoms ( C) by placing a 3 in front of the CO2 in the products.
C3H8 + O2 -----> 3 CO2 + H2O
3. Then balance the hydrogen atoms (H) by placing a 4 in front of the H2O in the products.
C3H8 + O2 -----> 3 CO2 + 4 H2O
4. Finally balance the the oxygen atoms (O) by placing a 5 in front of the O2 in the reactants.
C3H8 + 5 O2-----> 3 CO2 + 4 H2O
If the alkane contains an even number of carbon atoms, you will need to alter your strategy slightly.
You still use C H O; however when you attempt to balance the oxygens in the end you will get stuck.
A simple strategy for balancing alkanes that have an even number of hydrogens is to simply place a 2 in front of the alkane before you begin to balance.
Balancing combustion reactions with an even number of carbons in the alkane.
1) Let's first write the reaction.
C2H6 + O2 -----> CO2 + H2O
2) Arbitrarily place a 2 in front of the C2H6 in the reactants.
2 C2H6 + O2 -----> CO2 + H2O
3) Now balance the carbon atoms by placing a 4 in front of CO2 in the products.
2 C2H6 + O2 -----> 4 CO2 + H2O
4) Balance the hydrogen atoms by placing a 6 in front of H2O in the products.
2 C2H6 + O2 -----> 4 CO2 + 6 H2O
5) Now balance the oxygen atoms by placing a 7 in front of the O2 in the reactants.
2 C2H6 + 7 O2 -----> 4 CO2 + 6 H2O