# How would you balance: H2 + O2 --> H2O?

Dec 6, 2015

$2 {H}_{2} + {O}_{2} \rightarrow 2 {H}_{2} O$

#### Explanation:

${H}_{2} + {O}_{2} \rightarrow {H}_{2} O$

Notice the imbalance of atoms. On the left side, the reactant side, there are $2$ hydrogen atoms and $2$ oxygen atoms. On the right side, the product side, there are $2$ hydrogen atoms and only $1$ oxygen atom.

An important thing to remember when balancing equations is that the molecules themselves may not be changed—only their coefficients can. For example, we can change ${H}_{2} O$ to $3 {H}_{2} O$ but we can't go from ${H}_{2} O$ to ${H}_{3} O$.

In order to deal with the original imbalance, the $2$ oxygens on the left and $1$ on the right, we can change the coefficient of the ${H}_{2} O$ molecule from $1$ to $2$.

${H}_{2} + {O}_{2} \rightarrow 2 {H}_{2} O$

Now we have the same amount of oxygens on each side, $2$, but we have an unequal amount of hydrogen—there are $2$ on the left and $4$ on the right. We should achieve the least common multiple of the existing numbers, which is $4$. We can make the ${H}_{2}$ on the left have a coefficient of $2$, giving us the $4$ hydrogens that we need on that side of the equation, to match the $4$ on the right.

$2 {H}_{2} + {O}_{2} \rightarrow 2 {H}_{2} O$