Question

1. Correct the formulae and balance the following equations:
(1) K(s) + H2O(1) → KOH (aq) + H(g)
​

Answer 1

Explanation:

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

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.

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(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)#

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.

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:

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)

hope it's help you..