7.75% nitrogen in the sample.
The balanced reaction of HCl and NH3 is
HCl + NH3 = NH4Cl
Which means that 1 mole of HCl will neutralize 1 mole of HN3.
Since we used 83.0 ml of 0.150 m HCl, that means that we used
0.083 l * 0.150 mol/l = 0.01245 mol
So we had 0.01245 mol of NH3 which means that we had 0.01245 mol of N
Looking up the atomic weight of nitrogen, we get 14.0067. So
0.01245 * 14.0067 = 0.174383415 g
We now know that the sample had 0.17438 g of nitrogen. Divide by the original mass of the sample, getting
0.17438 / 2.25 = 0.077502222 = 7.7502222%
Rounding the result to 3 significant figures, gives us a mass percentage of 7.75% nitrogen in the original sample.
