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?

My answer is below. I am not sure if I am figuring it right. Please advise.

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?

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?

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.

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

Thank you!

Great explanation DrBob222!

-277

To calculate the oxidation state of nitrogen in HNO3, you need to consider the following rules:

1. The oxidation state of an element in its elemental form is always 0. In this case, oxygen (O2) has an oxidation state of 0.

2. The oxidation state of hydrogen (H) is usually +1, unless it combines with a metal, where it becomes -1.

3. The sum of the oxidation states of all the atoms in a compound should equal the charge on the compound. In this case, the compound is neutral, so the sum of the oxidation states should be 0.

Now, let's calculate the oxidation state of nitrogen in HNO3.

In ammonia (NH3), as you correctly mentioned, nitrogen (N) has an oxidation state of -3, and hydrogen (H) has an oxidation state of +1.

In nitric acid (HNO3), hydrogen (H) still has an oxidation state of +1, and oxygen (O) has an oxidation state of -2. Let's consider the oxidation state of nitrogen as X.

Using the rule that the sum of the oxidation states should equal 0, we can set up the equation:

(+1) + 3(-2) + X = 0

Simplifying the equation, we get:

+1 - 6 + X = 0

-5 + X = 0

X = +5

Therefore, the oxidation state of nitrogen in HNO3 is +5. Now, to determine if nitrogen is oxidized or reduced, you need to compare the oxidation state of nitrogen in the reactant (NH3) to the oxidation state in the product (HNO3).

In NH3, nitrogen has an oxidation state of -3, while in HNO3, it has an oxidation state of +5. Since the oxidation state has increased from -3 to +5, nitrogen has been oxidized.

Therefore, in the reaction NH3(g) + 2O2(g) → HNO3(aq) + H2O(l), nitrogen is oxidized.