Question

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

Answer 1

The coefficient go `H_2` is `1`.

Explanation:

`K(s)+color(blue)(H_2)O(l)->KOcolor(blue)(H)(aq)+color(blue)(H_2)(g)`

By looking at this equation, we can notice that we have `color(blue)(2H)` in the left side and `color(blue)(3H)` to the right side.

In order to balance this equation, we can multiply the `H_2` by `color(red)(1/2)`. Then we get:

`K(s)+color(blue)(H_2)O(l)->KOcolor(blue)(H)(aq)+color(red)(1/2)color(blue)(H_2)(g)`

This way, we will have `color(blue)(2H)` in each side.

Since we prefer to have the coefficients as whole numbers and not fractions, we can multiply the entire reaction by `color(green)(2)` so we get:

`color(green)(2)K(s)+color(green)(2)color(blue)(H_2)O(l)->color(green)(2)KOcolor(blue)(H)(aq)+color(blue)(H_2)(g)`