How do you balance the reaction #"NO" + "O"_2 -> "NO"_2#?

1 Answer
Truong-Son N.
Sep 30, 2015

Unbalanced:
#NO + O_2 -> NO_2#

Balancing:
Balance the nitrogen atoms first, because oxygen is by itself, and we can correct for the change in the number of oxygen atoms without affecting the number of nitrogen atoms. Instinctly I would do this because I see three oxygens on the left and two on the right:

#3NO + O_2 -> 3NO_2#

Realize that on the left side, we have an odd number of oxygens and on the right side, even. So, we have to double everything that we just touched.

#6NO + O_2 -> 6NO_2#

Currently, we have #6# nitrogens on both sides. Since on the right side we have #12# oxygens, and on the left we have #6# in #NO#, we need #6# more oxygens. Therefore:

#6NO + 3O_2 -> 6NO_2#

#N: [6] = [6]#
#O: [6 + 3*2 = 12] = [6*2 = 12]#

Since all three coefficients can be divided by #3#, do that:

#color(blue)(2NO + O_2 -> 2NO_2)#

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