How would you balance: KO2(s) + H2O(l) --? KOH(aq) + O2(g) +H2O2(aq)?

1 Answer
Nov 13, 2015

Answer:

Provided that the given reactants and products are correct, the balanced equation is #2KO_2# + #2H_2O# #rarr# #2KOH# + #O_2# + #H_2O_2#.

Explanation:

First tally all the atoms involved in the reaction. You can do this by looking at the subscripts.

#KO_2# + #H_2O# #rarr# #KOH# + #O_2# + #H_2O_2# (unbalanced)

left side:
K = 1
O = 2 + 1 (do not add this up yet)
H = 2

right side:
K = 1
O = 1 + 2 + 2 (do not add this up yet)
H = 1 + 2 (do not add this up yet)

Always remember that in balancing equations, you are not allowed to change the subscript, but you are allowed to put coefficients before the chemical formulas.

Let's start to balance the easiest atom, in this case the #H# atom. Since there are 2 #H# atoms on the left and 3 #H# atoms on the right (one coming from #KOH# and the other two coming from #H_2O_2#), you need to find a factor that you can multiply in either one of the substances to produce a desirable number to balance.

Since #KOH# is the simplest of the three substances with #H# atoms, we'll start the balancing there.

#KO_2# + #H_2O# #rarr# #color (red) 2KOH# + #O_2# + #H_2O_2#

left side:
K = 1
O = 2 + 1
H = 2

right side:
K = (1 x #color (red) 2#)
O = (1 x #color (red) 2#) + 2 + 2
H = (1 x #color (red) 2#) + 2

Notice that since the #H# atom is bonded to one #K# atom and one #O# atom, we also need to apply the coefficients to these elements. Now you have a total of 4 #H# atoms on the right, so to balance the #H# atoms on the left,

#KO_2# + #color (blue) 2H_2O# #rarr# #2KOH# + #O_2# + #H_2O_2#

left side:
K = 1
O = 2 + (1 x #color (blue) 2#)
H = 2 x #color (blue) 2# = 4

right side:
K = 1 x 2 = 2
O = (1 x 2) + 2 + 2
H = (1 x 2) + 2 = 4

Also, you have two #K# atoms on the right but only one #K# atom on the left. Thus,

#color (green) 2KO_2# + #2H_2O# #rarr# #2KOH# + #O_2# + #H_2O_2#

left side:
K = 1 x #color (green) 2# = 2
O = (2 x #color (green) 2#) + (1 x 2)
H = 2 x 2 = 4

right side:
K = 1 x 2 = 2
O = (1 x 2) + 2 + 2
H = (1 x 2) + 2 = 4

Now the only thing left to balance are the #O# atoms. But then again if you get the sum of #O# atoms on both sides of the equation,

#2KO_2# + #2H_2O# #rarr# #2KOH# + #O_2# + #H_2O_2#

left side:
K = 1 x 2 = 2
O = (2 x 2) + (1 x 2) = 6
H = 2 x 2 = 4

right side:
K = 1 x 2 = 2
O = (1 x 2) + 2 + 2 = 6
H = (1 x 2) + 2 = 4

The equation is already balanced.