17)

Sơ đồ phản ứng:

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

 Ta có:

\({n_{{O_2}}} = \frac{{28,8}}{{32}} = 0,9{\text{ mol}}\)

\({n_{C{O_2}}} = \frac{{39,6}}{{44}} = 0,9{\text{ mol;}}{{\text{n}}_{{H_2}O}} = \frac{{20,16}}{{22,4}} = 0,9{\text{ mol}}\)

Ta có:

\({n_C} = {n_{C{O_2}}} = 0,9{\text{ mol;}}{{\text{n}}_H} = 2{n_{{H_2}O}} = 1,8{\text{ mol}}\)

Bảo toàn \(O\)

\({n_{O{\text{ trong X}}}} + 2{n_{{O_2}}} = 2{n_{C{O_2}}} + {n_{{H_2}O}}\)

\( \to {n_{O{\text{ trong X}}}} + 0,9.2 = 0,9.2 + 0,9 \to {n_{O{\text{ trong X}}}} = 0,9{\text{ mol}}\)

Ta có:

\(C:H:O = 0,9:1,8:0,9 = 1:2:1\)

Vậy \(X\) có dạng \((CH_2O)_n\)

Ta có:

\(5,86.29 < {M_X} < 6,55.29 \to 169,94 < 30n < 189,95\)

Thỏa mãn \(n=6\)

Vậy \(X\) là \(C_6H_{12}O_6\)

\(m = {m_C} + {m_H} + {m_O} = 0,9.12 + 1,8.1 + 0,9.16 = 27{\text{ gam}}\)

18)

Ta có:

\({V_{C{H_4}}} = 33.6.70\%  = 23,52{\text{ lít}} \to {{\text{V}}_{{C_2}{H_6}}} = 10,08{\text{ lít}}\)

Đốt cháy hỗn hợp

\(C{H_4} + 2{O_2}\xrightarrow{{{t^o}}}C{O_2} + 2{H_2}O\)

\(2{C_2}{H_6} + 7{O_2}\xrightarrow{{{t^o}}}4C{O_2} + 6{H_2}O\)

\( \to {V_{{O_2}}} = 2{V_{C{H_4}}} + \frac{7}{2}{V_{{C_2}{H_6}}}\)

\( = 23,52.2 + 10,08.\frac{7}{2} = 82,32{\text{ lít}}\)

\( \to {V_{kk}} = 5{V_{{O_2}}} = 82,32.5 = 411,6{\text{ lít}}\)

\({n_{C{H_4}}} = \frac{{23,52}}{{22,4}} = 1,05{\text{ mol;}}{{\text{n}}_{{C_2}{H_6}}} = \frac{{10,08}}{{22,4}} = 0,45{\text{ mol}}\)

\( \to {n_{C{O_2}}} = {n_{C{H_4}}} + 2{n_{{C_2}{H_6}}} = 1,05 + 0,45.2 = 1,95{\text{ mol}}\)

\({n_{{H_2}O}} = 2{n_{C{O_2}}} + 3{n_{{C_2}{H_6}}} = 1,05.2 + 0,45.3 = 3,45{\text{ mol}}\)

\( \to {m_{C{O_2}}} = 1,95.44 = 85,8{\text{ gam}}\)

\({m_{{H_2}O}} = 3,45.18 = 62,1{\text{ gam}}\)

\({m_X} = {m_{C{H_4}}} + {m_{{C_2}{H_6}}} = 1,05.16 + 0,45.30 = 30,3{\text{ gam}}\)

\({n_X} = 1,05 + 0,45 = 1,5{\text{ mol}}\)

\( \to {M_X} = \frac{{30,3}}{{1,5}} = 20,2\)

\( \to {d_{X/kk}} = \frac{{20,2}}{{29}} = 0,6966\)