Đáp án:
\(A\)
Giải thích các bước giải:
Phản ứng xảy ra:
\({C_6}{H_{10}}{O_5} + 3HN{O_3}\xrightarrow{{{H_2}S{O_4}.{t^o}}}{C_6}{H_7}{O_2}{(N{O_3})_3} + 3{H_2}O\)
Ta có:
\({n_{{C_6}{H_7}{O_2}{{(N{O_3})}_3}}} = \frac{{59,4}}{{12.6 + 7 + 16.2 + 62.3}} = 0,2{\text{ kmol}}\)
\( \to {n_{HN{O_3}{\text{ lt}}}} = 3{n_{{C_6}{H_7}{O_2}{{(N{O_3})}_3}}} = 0,2.3 = 0,6{\text{ kmol}}\)
Vì hiệu suất là 90%.
\( \to {n_{HN{O_3}}} = \frac{{0,6}}{{90\% }} = \frac{2}{3}{\text{ kmol}}\)
\( \to {m_{HN{O_3}}} = \frac{2}{3}.63 = 42{\text{ kg}}\)
\( \to {m_{dd\;{\text{HN}}{{\text{O}}_3}}} = \frac{{42}}{{68\% }} = 61,765{\text{ kg}}\)
\( \to {V_{dd\;{\text{HN}}{{\text{O}}_3}}} = \frac{{61,765}}{{1,52}} = 40,63{\text{ lít}}\)