# Chemistry

NH3(g) + 2O2 (g) = HNO3 (aq) + H2O (l)

How do I calculate the oxidation states of Nitrogen and figure out if it is reduced, oxidized or neither?

NH3(g) + 2O2 (g) = HNO3 (aq) + H2O (l)
N is -3,H3 is +1, 2o2 is 0,

H is +1, N is X ??, O3 is -2, H2O is -2

Therefore X=-2
I am not sure how to figure N. Is it X or is N include NO3?
and it is

Is it Oxidized?

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1. You seem to be confused over this.
You have N = -3 which is correct in NH3. So why X? What are you solving for and why if you know N is -3?

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2. That is what I am confused about. Do you separate the N out of the HNO3? So, it is H+1, N -3, O3 is -2? I guess I am making harder then it is.

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3. right. Have you looked at this site to help do oxidation states. Here is a quick primer, then the site.
For NH3. H is +1, total for H is +3; therefore, N is -3. For the right side. HNO3. H is +1, oxygen is -2, total oxygen is -6, so what must N be to make HNO3 zero (remember all compounds are zero). Obviously 1 + 5 +(-6) = 0 so N in HNO3 is +5. Do you separate NO3? It really makes no difference. I prefer to deal with molecules BUT if you wish to separate into NO3^-, then O is -2, 3x-2=-6 and we say what must N be to leave a -1 charge. Obviously +5-6= -1.
O on the left is zero. O in H2O is -2.

What's oxidized. Just remember the definition. Leo the lion goes grrr. Leo===loss electrons oxidation. Which lost electrons?

http://www.chemteam.info/Redox/Redox-Rules.html

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4. Thank you!

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5. Great explanation DrBob222!

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6. -277

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