1)
\({C_6}{H_{10}}{O_5} + 3HN{O_3}\xrightarrow{{{H_2}S{O_4},{t^o}}}{C_6}{H_7}{O_2}{(N{O_3})_3} + 3{H_2}O\)
Ta có:
\({n_{{C_6}{H_{10}}{O_5}}} = \frac{{1,62}}{{12.6 + 10 + 16.5}} = 0,01 = {n_{{C_6}{H_7}{O_2}{{(N{O_3})}_3}{\text{ lt}}}}\)
Vì hao hụt 12%.
\( \to {n_{{C_6}{H_7}{O_2}{{(N{O_3})}_3}}} = 0,01.(100\% - 12\% ) = 0,0088\)
\( \to {m_{{C_6}{H_7}{O_2}{{(N{O_3})}_3}}} = 0,0088.(12.6 + 7 + 16.2 + 62.3) = 2,6136{\text{ tấn}}\)
2)
\({C_6}{H_{10}}{O_5} + 3HN{O_3}\xrightarrow{{{H_2}S{O_4},{t^o}}}{C_6}{H_7}{O_2}{(N{O_3})_3} + 3{H_2}O\)
\( \to {n_{{C_6}{H_7}{O_2}{{(N{O_3})}_3}}} = \frac{{59,4}}{{12.6 + 7 + 16.2 + 62.3}} = 0,2\)
\( \to {n_{HN{O_3}{\text{ lt}}}} = 3{n_{{C_6}{H_7}{O_2}{{(N{O_3})}_3}}} = 0,2.3 = 0,6\)
Vì hiệu suất là 90%
\( \to {n_{HN{O_3}}} = \frac{{0,6}}{{90\% }} \to {m_{HN{O_3}}} = \frac{{0,6}}{{90\% }}.63 = 42{\text{ kg}}\)
\( \to {m_{dd{\text{ HN}}{{\text{O}}_3}}} = \frac{{42}}{{99,67\% }} = 42,139{\text{ kg}}\)
\( \to {V_{dd{\text{ HN}}{{\text{O}}_3}}} = \frac{{42,139}}{{1,52}} = 27,723{\text{ lít}}\)
