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

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Answer 1 · 2016-08-19T08:36:11

The coefficient go #H_2# is #1#.

#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)#