# Why are combustion reactions taught as BOTH word equations, and symbol equations?

##### 1 Answer
Oct 12, 2016

Why, because teachers and educators have found this to be the most efficient, most intuitive method.

#### Explanation:

I take it (and perhaps I am wrong), that you speak of combustion reactions, where a hydrocarbon is completely (or incompletely) combusted with dioxygen to give carbon dioxide and water. So for pentane, we write:

$\text{Pentane + oxygen "rarr" carbon dioxide + water}$

If we wished to do this symbolically, and we do!, we write the unbalanced equation:

${C}_{5} {H}_{12} + {O}_{2} \rightarrow C {O}_{2} + {H}_{2} O$

So balance the carbon:

${C}_{5} {H}_{12} + {O}_{2} \rightarrow 5 C {O}_{2} + {H}_{2} O$

And the the hydrogen:

${C}_{5} {H}_{12} + {O}_{2} \rightarrow 5 C {O}_{2} + 6 {H}_{2} O$

And then the oxygen:

${C}_{5} {H}_{12} + 8 {O}_{2} \rightarrow 5 C {O}_{2} + 6 {H}_{2} O$

And thus we have balanced in order, (i) the carbons, and (ii) then the hydrogens, and (iii) then the oxygens to give the stoichiometrically balanced equation. When you do this for even numbered alkanes, sometimes this is problematic.

${C}_{6} {H}_{14} + \frac{19}{2} {O}_{2} \rightarrow 6 C {O}_{2} + 7 {H}_{2} O$

Some would maintain that you cannot have half integral coefficients, and instead you should double the equation to give:

$2 {C}_{6} {H}_{14} + 19 {O}_{2} \rightarrow 12 C {O}_{2} + 14 {H}_{2} O$

I prefer the first equation with its half integral coefficient, because I find that the arithmetic is easier to rationalize and intuit in a subsequent calculation where given masses are reacted. Whichever method you choose, both mass and charge must be conserved, and $\text{GARBAGE IN "-=" GARBAGE OUT}$.