\(25 \times N = 30 \times 0.1 \times 2\)
\({N_{HCl}} = \frac{{30 \times 0.2}}{{25}} = \frac{6}{5} \times 0.2 = \frac{{1.2}}{5}\)
For the \(2^{nd}\) titration
\(\frac{{1.2}}{5} \times {V_{HCl}} = 30 \times 0.2\)
\({V_{HCl}} = \frac{{6 \times 5}}{{1.2}} = \frac{{30}}{{1.2}} = 25\,ml\)