Is `CH_4 + 2O_2 = CO_2 + 2H_2O` a balanced equation? It seems to have too many oxygen atoms on the right of the equation to me.

Wednesday, January 22, 2014

Is $\displaystyle{C}{H}_{{4}}+{2}{O}_{{2}}={C}{O}_{{2}}+{2}{H}_{{2}}{O}$ a balanced equation? It seems to have too many oxygen atoms on the right of the equation to me.

This equation seems completely balanced to me. Let's figure out if this is true.

For an equation to be a balanced one, the quantity of atoms of each type must be the same on the left side and on the right side. Do not forget that an index $\displaystyle{n}$ under atom in a formula means that this molecule has $\displaystyle{n}$ atoms of that type.

There are three types of atoms in our reaction: $\displaystyle{C},$ $\displaystyle{H},$ and $\displaystyle{O}.$ Let's start from $\displaystyle{O}.$

There are $\displaystyle{0}$ oxygen atoms in $\displaystyle{C}{H}_{{4}},$ $\displaystyle{2}\cdot{2}$ oxygen atoms in two molecules of $\displaystyle{O}_{{2}},$ $\displaystyle{2}$ in $\displaystyle{C}{O}_{{2}}$ and $\displaystyle{2}$ in two molecules of $\displaystyle{H}_{{2}}{O}.$ So, we have $\displaystyle{0}+{2}\cdot{2}={4}$ oxygen molecules at the left side and $\displaystyle{2}+{2}={4}$ at the right one. Balanced.

For $\displaystyle{C},$ we have $\displaystyle{1}+{0}={1}+{0},$ also true. For $\displaystyle{H},$ $\displaystyle{4}+{0}={0}+{2}\cdot{2},$ true. As a whole, the equation is balanced.