Numerical answer type, answer in %, tolerance 0.02.
Answer: 5.01 %
H+ supplied = 2 x 0.0400 x 0.1000 = 0.0080 mol. NaOH used = 0.0221 x 0.2000 = 0.00442 mol = H+ left over. H+ neutralised by NH3 = 0.00358 mol = mol N. Mass N = 0.00358 x 14 = 0.0501 g; percent = 5.012, about 5.01%.