Question 1First, recognise that after the preparatory reaction (the given reaction), we are left with a base: NH3.
Strategy:- Calculate the number of mole of hydrochloric acid consumed per aliquot.
- Realise there will be no excess

, hence ammonia will react with hydrochloric acid.
- Write an equation for the reaction between hydrochloric acid and ammonia (acid + base).
- Hence, find the number of mole of ammonia that is present per aliquot.
- Using this, calculate the number of mole of ammonium present (using the given reaction) per aliquot.
- Scale this number to the number of moles of ammonium present in the 1.3 gram sample.
- Hence find the mass of
nitrogen (N only), and the percentage mass of
nitrogenSolution:} = 0.0443\times 0.1 = 0.00443\mbox{ mol})
There is no excess

, because it is all neutralised with the ammonium, so the reaction that takes place is:
Chemical equation:

}_{\mbox{aliquot}} = \mbox{n(NH3)} = \mbox{n(HCl)} = 0.00443\mbox{ mol})
}_{\mbox{sample}} = 0.00443 \times \frac{250}{20} = 0.0554\mbox{ mol})
} = \mbox{n(NH4}^+\mbox{)} \times \mbox{M(N)} = 0.0554 \times 14 = 0.775\mbox{ g})
})
(It should be 1 significant figure, but since that's ridiculous, I've given it to three)