NH3(aq)+HCl(aq)----->NH4Cl(aq)
If 46.0 mL of 0.150 M HCl(aq) is needed to neutralize all the NH3(g) from a 2.25-g sample of organic material, calculate the mass percentage of nitrogen in the sample.
(0.0460 L) x (0.150 mol /L HCl) x (1 mol NH3 / 1 mol HCl) x
(1 mol N / 1 mol NH3) x (14.00672 g N / mol) / (2.25 g) = 0.0430 = 4.30% N