A level chemistry question

So guys the question is....

Under some conditions, nitrate(V) ions (NO3–) can oxidise Mn2+ ions to MnO2 and the nitrate(V) ions are reduced.
An experiment is used to find the oxidation state of nitrogen at the end of the reaction.
In this experiment, 24.0 cm3 of 0.150 mol dm–3 aqueous solution of Mn2+ ions react with exactly 15.0 cm3 of 0.0960 mol dm–3 aqueous solution of NO3– ions.
Write a half-equation for the oxidation of Mn2+ to MnO2
Calculate the simplest whole-number reacting ratio of Mn2+ ions to NO3– ions.
Determine the oxidation state of N at the end of the reaction.
[5 marks]

So i got 4/5

my answers:
half equation = Mn^2+ + 2 H2O → MnO2 + 4 H+ + 2 e–

moles of Mn^2+ = 3.60 x 10^-3
moles of NO3- = 1.44 x 10^-3

ratio Mn2+: NO3-

5:3

but the answer for the oxidation state of N is 0. Why is that? How do i find that out?

Not entirely sure, but if the final oxidation state is zero, then the product from the redox reaction must be nitrogen gas. Is there a way to prove that using the ratios?

So guys the question is....

Under some conditions, nitrate(V) ions (NO3–) can oxidise Mn2+ ions to MnO2 and the nitrate(V) ions are reduced.
An experiment is used to find the oxidation state of nitrogen at the end of the reaction.
In this experiment, 24.0 cm3 of 0.150 mol dm–3 aqueous solution of Mn2+ ions react with exactly 15.0 cm3 of 0.0960 mol dm–3 aqueous solution of NO3– ions.
Write a half-equation for the oxidation of Mn2+ to MnO2
Calculate the simplest whole-number reacting ratio of Mn2+ ions to NO3– ions.
Determine the oxidation state of N at the end of the reaction.
[5 marks]

So i got 4/5

my answers:
half equation = Mn^2+ + 2 H2O → MnO2 + 4 H+ + 2 e–

moles of Mn^2+ = 3.60 x 10^-3
moles of NO3- = 1.44 x 10^-3

ratio Mn2+: NO3-

5:3

but the answer for the oxidation state of N is 0. Why is that? How do i find that out?