1)

Phản ứng xảy ra:

\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)

Ta có: 

\({n_{Fe}} = \frac{{11,2}}{{56}} = 0,2{\text{ mol}} \to {{\text{n}}_{HCl}} = 2{n_{Fe}} = 0,4{\text{ mol}} \to {\text{a = }}\frac{{0,4}}{{0,2}} = 2M\)

Ta có: 

\({n_{FeC{l_2}}} = {n_{Fe}} = {n_{{H_2}}} = 0,2{\text{ mol}} \to {\text{m = 0}}{\text{,2}}{\text{.(56 + 35}}{\text{,5}}{\text{.2) = 25}}{\text{,4 gam; V = 0}}{\text{,2}}{\text{.22}}{\text{,4 = 4}}{\text{,48 lít}}\)

2)

Phản ứng xảy ra:

\(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)

Ta có: 

\({n_{Mg}} = \frac{{1,44}}{{24}} = 0,06{\text{ mol = }}{{\text{n}}_{{H_2}}} \to V = 0,06.22,4 = 1,344{\text{ lít}}\)

\({n_{HCl}} = 2{n_{Mg}} = 0,12{\text{ mol}} \to {{\text{m}}_{HCl}} = 0,12.36,5 = 4,38{\text{ gam}} \to {{\text{m}}_{dd{\text{ HCl}}}} = \frac{{4,38}}{{3,65\% }} = 120gam\)

3)

Gọi n là hóa trị của R

Phản ứng xảy ra:

\(R + nHCl\xrightarrow{{}}RC{l_n} + \frac{n}{2}{H_2}\)

\({n_{{H_2}}} = \frac{{10,08}}{{22,4}} = 0,45{\text{ mol}} \to {{\text{n}}_R} = \frac{{2{n_{{H_2}}}}}{n} = \frac{{0,9}}{n} \to {M_R} = \frac{{25,2}}{{\frac{{0,9}}{n}}} = 28n \to n = 2;R = 56 \to Fe\)