In these kinds of problems start from the ending. Here equivalents present in NaOH = 80* 0.5 = 40 m.eqns hence the total equivalents ofH2 SO4used form NaOH = 40 m.eqns total equivalents ofH 2 SO 4. = 50*0.5*2....(here 2 for giving two hydrogens per molecule, it is also called n-factor) = 50 m.eqns hence the number of equivalents ammonia = 50 - 40 = 10 m.eqns. 1 equivalent contains 1 nitrogen molecule, hence total weight of nitrogen = 10* 10-3 * 14 = 0.14 grams hence percentage of nitrogen = 0.14/0.7 *100 = 20 %