Having trouble balancing a chemical equation. How would I balance this?

Question

#NH_3# + #O_2# #rarr# NO + #H_2O#

I find that I have 1 N on both sides, 3 H's on the left and 2 H's on the right. The O's are balanced. I can balance the H's (giving them a total of 6 and both sides), but I can't seem to figure out how to do with the others, without throwing the whole thing off.

Mateo Aguilar · Accepted Answer

Here's how you can go about balancing this chemical equation.

If, as is assumed here, you do not wish to use oxidation numbers, this chemical equation can be a little difficult to balance. When faced with such situations, the first thing to do is to make sure that the elements that are present in odd numbers on one side of the equation and in even numbers on the other are balanced. In this case, you have #3# atoms of hydrogen on the reactants' side and #2# on the products' side. To balance these out, multiply the ammonia molecule by #2# and the water molecule by #3# #color(red)(2)"NH"_3 + "O"_2 -> "NO" + color(green)(3)"H"_2"O"# The atoms of nitrogen and oxygen are now out of balance; pay attention to the nitrogen atoms first and the oxygen atoms last; you'll understand why in a moment. Since you now have #2# atoms of nitrogen on the reactants' side, and only #1# on the products' side, multiply the nitric oxide molecule by #2# to get #color(red)(2)"NH"_3 + "O"_2 -> color(blue)(2)"NO" + color(green)(3)"H"_2"O"# Now everything is balanced except the oxygen atoms, since you have #5# on the products' side and only #2# on the reactants' side. On the reactants' side, oxygen is present as a molecule, #"O"_2#, which means that you have some elbow room in terms of what stoichiometric coefficient you can assign here. So, what number multiplied by #2# will give you #5#? #x * 2= 5 implies x= 5/2# This means that if you multiply the oxygen molecule by #5/2#, you will have #5# oxygen atoms on both sides of the equation. #color(red)(2)"NH"_3 + 5/2"O"_2 -> color(blue)(2)"NO" + color(green)(3)"H"_2"O"# Finally, to remove the fractional coefficient, multiply all the species by #2# #color(red)(4)"NH"_3 + 5"O"_2 -> color(blue)(4)"NO" + color(green)(6)"H"_2"O"# Here is the chemical equation for this reaction that is in balance.

Owen Ambrose · Answer

undefined Sure, please provide the chemical equation you're having trouble balancing, and I'll assist you in balancing it.

Having trouble balancing a chemical equation. How would I balance this?

#NH_3# + #O_2# #rarr# NO + #H_2O# I find that I have 1 N on both sides, 3 H's on the left and 2 H's on the right. The O's are balanced. I can balance the H's (giving them a total of 6 and both sides), but I can't seem to figure out how to do with the others, without throwing the whole thing off.

#NH_3# + #O_2# #rarr# NO + #H_2O#

I find that I have 1 N on both sides, 3 H's on the left and 2 H's on the right. The O's are balanced. I can balance the H's (giving them a total of 6 and both sides), but I can't seem to figure out how to do with the others, without throwing the whole thing off.