1.

Sơ đồ phản ứng:

\(X + {O_2}\xrightarrow{{}}C{O_2} + {H_2}O\)

Ta có: \({n_{C{O_2}}} = {n_C} = 0,4{\text{ mol;}}{{\text{n}}_{{H_2}O}} = \frac{9}{{18}} = 0,5{\text{ mol}} \to {{\text{n}}_H} = 2{n_{{H_2}O}} = 1{\text{ mol}}\)

Bảo toàn O: 

\({n_{O{\text{ trong X}}}} = 2{n_{C{O_2}}} + {n_{{H_2}O}} - 2{n_{{O_2}}} = 0,4.2 + 0,5 - 0,65.2 = 0\)

\({n_C}:{n_H} = 0,4:1 = 4:10 \to X:{({C_4}{H_{10}})_n} \to 58n = 58 \to n = 1\)

Vậy X là \({C_4}{H_{10}}\)

2)

\({( - {C_6}{H_{10}}{O_5} - )_n} + n{H_2}O\xrightarrow{{axit,{t^o}}}n{C_6}{H_{12}}{O_6}\)

Ta có: 

\({n_{{C_6}{H_{10}}{O_5}}} = \frac{1}{{162}} \to {n_{{C_6}{H_{10}}{O_5}{\text{ phản ứng}}}} = {n_{{C_6}{H_{12}}{O_6}}} = \frac{1}{{162}}.90\%  = \frac{1}{{180}} \to {m_{{C_6}{H_{12}}{O_6}}} = \frac{1}{{180}}.180 = 1{\text{ tấn}}\)