4)

Sơ đồ phản ứng:

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

a)

 Ta có: \({n_{C{O_2}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol < }}{{\text{n}}_{{H_2}O}} = \frac{{5,4}}{{18}} = 0,3{\text{ mol}}\) do vậy X là ankan.

\( \to {n_X} = {n_{{H_2}O}} - {n_{C{O_2}}} = 0,3 - 0,2 = 0,1{\text{ mol}} \to {{\text{C}}_X} = \frac{{0,2}}{{0,1}} = 2\)

Vậy X là \({C_2}{H_6}\)

b)

Ta có: \({n_{C{O_2}}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol < }}{{\text{n}}_{{H_2}O}} = \frac{{7,2}}{{18}} = 0,4{\text{ mol}}\)

\( \to {n_X} = {n_{{H_2}O}} - {n_{C{O_2}}} = 0,4 - 0,3 = 0,1{\text{ mol}} \to {{\text{C}}_X} = \frac{{{n_{C{O_2}}}}}{{{n_X}}} = \frac{{0,3}}{{0,1}} = 3 \to X:{C_3}{H_8}\)

11)

Phản ứng xảy ra:

\({C_6}{H_5} - C{H_3} + 3HN{O_3}\xrightarrow{{{H_2}S{O_4},{t^o}}}C{H_3} - {C_6}{H_2} - {(N{O_2})_3} + 3{H_2}O\)

\( \to {n_{TNT}} = {n_{toluen}} = \frac{{68,1}}{{15 + (12.6 + 2) + 46.3}} = 0,3{\text{ mol}} \to {\text{m = }}{{\text{m}}_{toluen}} = 0,3.(12.7 + 8) = 27,6{\text{ gam}}\)