1)

Sơ đồ phản ứng:

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

Ta có: 

\({n_{C{O_2}}} = \frac{{13,44}}{{22,4}} = 0,6{\text{ mol;}}{{\text{n}}_C};{n_{{H_2}O}} = \frac{{14,4}}{{18}} = 0,8{\text{ mol}} \to {{\text{n}}_H} = 2{n_{{H_2}O}} = 1,6{\text{ mol}} \to {{\text{n}}_O} = \frac{{12 - 0,6.12 - 1,6.1}}{{16}} = 0,2{\text{ mol}}\)

\( \to {n_C}:{n_H}:{n_O} = 0,6:1,6:0,2 = 3:8:1\)

X có dạng \({({C_3}{H_8}O)_n}\)

\({n_{{O_2}}} = \frac{{0,96}}{{32}} = 0,03{\text{ mol = }}{{\text{n}}_X} \to {M_X} = \frac{{1,8}}{{0,03}} = 60 = 60n \to n = 1 \to X:{C_3}{H_8}O\)

2)

Mỗi phần chứa 0,5m gam hỗn hợp.

Cho phần 1 tác dụng với axit.

\(Mg + {H_2}S{O_4}\xrightarrow{{}}MgS{O_4} + {H_2}\)

\(2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)

\(Zn + {H_2}S{O_4}\xrightarrow{{}}ZnS{O_4} + {H_2}\)

Ta có: 

\({n_{{H_2}}} = {n_{Mg}} + 1,5{n_{Al}} + {n_{Zn}} = \frac{{5,6}}{{22,4}} = 0,25{\text{ mol}}\)

Cho phần 2 tác dụng với oxi

\(2Mg + {O_2}\xrightarrow{{}}2MgO\)

\(4Al + 3{O_2}\xrightarrow{{{t^o}}}2A{l_2}{O_3}\)

\(2Zn + {O_2}\xrightarrow{{{t^o}}}2ZnO\)

Ta có: 

\({n_{{O_2}}} = \frac{1}{2}{n_{Mg}} + \frac{3}{4}{n_{Al}} + \frac{1}{2}{n_{Zn}} = \frac{{0,25}}{2} = 0,125{\text{ mol}}\)

BTKL: 

\(0,5m + {m_{{O_2}}} = {m_{oxit}} \to m = 11,15 - 0,125.32 = 7,15{\text{gam}} \to {\text{m = 14}}{\text{,3 gam}}\)