2)

M hóa trị III mới đúng nha em.

Phản ứng xảy ra:

\(2M + 6HCl\xrightarrow{{}}2MC{l_3} + 3{H_2}\)

Ta có: 

\({n_{{H_2}}} = \frac{{10,08}}{{22,4}} = 0,45{\text{ mol}} \to {{\text{n}}_M} = \frac{2}{3}{n_{{H_2}}} = 0,3{\text{ mol}} \to {{\text{M}}_M} = \frac{{8,1}}{{0,3}} = 27 \to M:Al\)

3)

Phản ứng xảy ra:

\(Ca + 2{H_2}O\xrightarrow{{}}Ca{(OH)_2} + {H_2}\)

Ta có: 

\({n_{Ca}} = {n_{Ca{{(OH)}_2}}} = {n_{{H_2}}} = \frac{4}{{40}} = 0,1{\text{ mol}}\)

\( \to m = {m_{Ca{{(OH)}_2}}} = 0,1.(40 + 17.2) = 7,4{\text{ gam; V = }}{{\text{V}}_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)

\(CuO + {H_2}\xrightarrow{{{t^o}}}Cu + {H_2}O\)

Ta có: 

\({n_{Cu}} = {n_{{H_2}}} = 0,1{\text{ mol}} \to {{\text{m}}_{Cu}} = 0,1.64 = 6,4{\text{ gam}}\)