# When the following equation is balanced, K(s) + H_2O(l) -> KOH(aq) + H_2(g), what is the coefficient of H_2?

Aug 19, 2016

The coefficient go ${H}_{2}$ is $1$.

#### Explanation:

$K \left(s\right) + \textcolor{b l u e}{{H}_{2}} O \left(l\right) \to K O \textcolor{b l u e}{H} \left(a q\right) + \textcolor{b l u e}{{H}_{2}} \left(g\right)$

By looking at this equation, we can notice that we have $\textcolor{b l u e}{2 H}$ in the left side and $\textcolor{b l u e}{3 H}$ to the right side.

In order to balance this equation, we can multiply the ${H}_{2}$ by $\textcolor{red}{\frac{1}{2}}$. Then we get:

$K \left(s\right) + \textcolor{b l u e}{{H}_{2}} O \left(l\right) \to K O \textcolor{b l u e}{H} \left(a q\right) + \textcolor{red}{\frac{1}{2}} \textcolor{b l u e}{{H}_{2}} \left(g\right)$

This way, we will have $\textcolor{b l u e}{2 H}$ in each side.

Since we prefer to have the coefficients as whole numbers and not fractions, we can multiply the entire reaction by $\textcolor{g r e e n}{2}$ so we get:

$\textcolor{g r e e n}{2} K \left(s\right) + \textcolor{g r e e n}{2} \textcolor{b l u e}{{H}_{2}} O \left(l\right) \to \textcolor{g r e e n}{2} K O \textcolor{b l u e}{H} \left(a q\right) + \textcolor{b l u e}{{H}_{2}} \left(g\right)$