Balancing chemical equation: Propance `(C_(3) H_(8))` is a colorless, odorless gas often used as a heating and cooking fuel in rural homes, Write a balanced equation for the combusion reaction of propane with oxygen to yield carbon dioxide and water.

Explanation

Stragey and solution:
Just follow the four steps described below:
Step1. Write down the unbalanced chemical equation using the correct chemical formulas for all substances, i.e., the reactions and products:
`C_(3) H_(8) (g) + O_(2) (g) rarr CO_(2) (g) + H_(2) O(I)`
This equation is not balanced because the coefficients that gives the relative number of molecules (formula unis) involved have not yet been determined. To balanced the equation, we need to select coefficients that will make the numbers of atoms to each element equal on both sides of the equcation.
Step2: Look at the unbalanced equation and note the following:
`underset(3C,8H,2O)("In reactant")" "underset(1C,2H,3O)("In products")`
Thus, `C, H` and `O` are not balanced.
To select the right cofficients to balance the equation, it is usually the best to start with the most complex substance (in this case `C_(3) H_(8)`) and to deal with one 3 element at a time3. There are `3 C` atoms on the left side but only 1 on the right side. Thus we write 3 for the coefficient of `CO_(2)` to balance the carbon atoms.

`{:(C_(3)H_(8)+O_(2),rarr,3CO_(2)+H_(2)O,),("In reactants",,"In products",),("3 C,8 H,2 O",,"3 C,2 H,7 O",("Balanced for C")):}`
There are 8 hydrogen atoms on the left (in `C(3)H_(8)`) but only 2 (in `H_(2)O`) on the right. Hence, we write 4 for the coefficient of `H_(2)O` to balance the hydrogen atoms :
`{:(C_(3)H_(8)+O_(2),rarr,3CO_(2)+H_(2)O,),("In reactants",,"In products",),("3 C,8 H,2 O",,"3 C,8 H,10 O",("Balanced for C and H")):}`
Finally, we determine the right coefficient for oxygen. There are 2 on the left (in `O_(2)`) but 10 on the right (6 from `3CO_(2)` and 4 from `4H_(2) O`). Hence to balanced the oxygen atoms, we wirte 5 for the coefficient of `O_(2)`:
`{:(C_(3)H_(8)+5O_(2),rarr,3CO_(2)+4H_(2)O,),("In reactants",,"In products",),("3 C,8 H,10 O",,"3 C,8 H,10 O",):}`
Now the equacation is balanced.
Step3: When we think we have completed the balancing we should always do a complete check for each element by counting its number values. In fact, our answer is already correct but the combusion of propane can be equallity well represented by
`2C_(3) H_(8) + 10 O_(2) rarr CO_(2) + 8H_(2) O`
In which the cofficients (arrived diffrently through trail-and-error) of all substances have been doubled. Usually it is preferable to write the coefficients so that they are the smallest whole numbers possible. Thus, when we refer to a balanced chemical equation for the combustion reaction of propane, we will assume that this is the equation we got in step 2 unless otherwise specified.