Đáp án:
\({V_{dd{\text{ HN}}{{\text{O}}_3}}} = 657,895{\text{ ml}}\)
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:
\({n_{{C_6}{H_7}{O_2}{{(N{O_2})}_3}}} = \frac{{594}}{{12.6 + 7 + 16.2 + 62.3}} = 2{\text{ mol}}\)
\( \to {n_{HN{O_3}{\text{ lt}}}} = 3{n_{{C_6}{H_7}{O_2}{{(N{O_3})}_3}}} = 2.3 = 6{\text{ mol}}\)
Vì hiệu suất là 60%
\( \to {n_{HN{O_3}}} = \frac{6}{{60\% }} = 10{\text{ mol}}\)
\( \to {m_{HN{O_3}}} = 10.63 = 630{\text{ gam}}\)
\( \to {m_{dd{\text{ HN}}{{\text{O}}_3}}} = \frac{{630}}{{63\% }} = 1000{\text{ gam}}\)
\( \to {V_{dd{\text{ HN}}{{\text{O}}_3}}} = \frac{{1000}}{{1,52}} = 657,895{\text{ ml}}\)