Im double checking my answer to a chemistry question, Ill but the question itself on details?

In the question, ammonium nitrate in Nitrochalk reacts with NaOH to give ammonia gas, but calcium carbonate does not. The excess NaOH in 25 cm³ portion (total 250 cm³) is titrated against HCl.

Consider the titration of excess NaOH in 25 cm³-portion against HCl.
NaOH + HCl → NaCl+ H₂O
Mole ratio NaOH : HCl = 1 : 1

No. of moles of HCl reacted = (0.100 mol/dm³) × (30.0/1000 dm³) = 0.003 mol
No. of moles of excess NaOH in the 25.0 cm³ portion = 0.003 mol
No. of moles of excess NaOH in 250 cm³ portion = (0.003 mol) × (250/25.0) = 0.03 mol

Total number of moles of NaOH used = (2.00 mol/dm³) × (25.0/1000 dm³) = 0.05 mol
No. of moles of NaOH reacted with NH₄NO₃ = (0.05 - 0.03) mol = 0.02 mol

Consider the reaction between NaOH and NH₄NO₃ in the nitrochalk :
NH₄NO₃ + NaOH → NaNO₃ + NH₃ + H₂O
Mole ratio NH₄NO₃ : NaOH = 1 : 1

No. of moles of NaOH reacted with NH₄NO₃ = 0.02 mol
No. of moles of NH₄NO₃ in Nitrochalk = 0.02 mol

Each mole of NH₄NO₃ contains 2 moles of N.
No. of moles of N in Nitrochalk = (0.02 mol) × 2 = 0.04 mol

Molar mass of N = 14.0 g/mol
Mass of NH₄NO₃ in Nitrochalk = (0.04 mol) × (14.0 g/mol) = 0.56 g

% by mass of N in Nitrochalk = (0.56/2.00) × 100(%) = 28.0(%)

Calculate mol NaOH used in reaction
Mol NaOH in 25.0cm³ of 2.00mol/dm³ solution = 25/1000*2.00 = 0.05 mol NaOH

Calculate mol of NaOH unreacted:
Equation:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
Mol HCl in 30.0cm³ of 0.100mol/dm³ solution = 30/1000*0.100 = 0.003 mol NaOH unreacted in the 25.0cm³
Total mol NaOH unreacted in the 250cm³ volume = 0.003*250/25 = 0.03 mol

Mol NaOH used to react with the NH4NO3 = 0.05 0.05 = 0.02 mol
Balanced equation:
NH4NO3(aq) + NaOH(aq) = NH3(g) + NaNO3(aq) + H2O(l)
1mol NH4NO3 reacts with 1 mol NaOH
Mol NH4NO3 present in sample = 0.02 mol
1mol NH4NO3 contains 2 mol N
0.02 mol NH4NO3 contains 0.04 mol N
Molar mass N = 14g/mol
Mass of 0.04 mol N = 0.04*14 = 0.56g N
% N = 0.56g/2.00g *100 = 28% N in Nitrochalk.
Does this answer seem correct?
Here is a paste from a description of ICI Nitrochalk:
The finished product is an easy and pleasant to handle pellet containing 26% nitrogen, half in the ammonical form and half as nitrate, which has nitrogen immediately available and some in reserve.
Seems OK