1)

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:

\({n_{{C_6}{H_{10}}{O_5}}} = \dfrac{{1,62}}{{12.6 + 10 + 16.5}} = 0,01 = {n_{{C_6}{H_7}{O_2}{{(N{O_3})}_3}{\text{ lt}}}}\)

Vì bị hao hụt là 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)

Phản ứng xảy ra:

\({(C{H_2}OH)_5}CHO + 2AgN{O_3} + 3N{H_3} + {H_2}O\xrightarrow{{}}2Ag + {(C{H_2}OH)_5}COON{H_4} + 2N{H_4}N{O_3}\)

Ta có:

\({n_{{C_6}{H_{12}}{O_6}}} = \dfrac{{27}}{{12.6 + 12 + 16.6}} = 0,15{\text{ mol}}\)

\( \to {n_{Ag{\text{ lt}}}} = 2{n_{{C_6}{H_{12}}{O_6}}} = 0,15.2 = 0,3{\text{ mol}}\)

Vì hiệu suất là $75\%$ nên

\({n_{Ag}} = 0,3.75\%  = 0,225{\text{ mol}}\)

\( \to {m_{Ag}} = 0,225.108 = 24,3{\text{ gam}}\)

3)

Phản ứng xảy ra:

\({C_6}{H_{12}}{O_6}\xrightarrow{{men}}2{C_2}{H_5}OH + 2C{O_2}\)

Ta có:

\({m_{{C_6}{H_{12}}{O_6}}} = 2,25.(100\%  - 20\% ) = 1,8{\text{ kg}}\)

\( \to {n_{{C_6}{H_{12}}{O_6}}} = \dfrac{{1,8}}{{12.6 + 12 + 16.6}} = 0,01{\text{ kmol}}\)

\( \to {n_{{C_2}{H_5}OH{\text{ lt}}}} = 2{n_{{C_6}{H_{12}}{O_6}}} = 0,01.2 = 0,02{\text{ kmol}}\)

Vì ancol bị hao hụt $10\%.$

\( \to {n_{{C_2}{H_5}OH}} = 0,02.(100\%  - 10\% ) = 0,018{\text{ kmol}}\)

\( \to {m_{{C_2}{H_5}OH}} = 0,018.46 = 0,828{\text{ kg}}\)