1)

Phản ứng xảy ra:

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

Ta có:

\({n_{Mg}} = \frac{8}{{24}} = \frac{1}{3}{\text{ mol > }}{{\text{n}}_{{H_2}S{O_4}}} = 0,2.1 = 0,2{\text{ mol}}\)

Vậy \(Mg\) dư

\( \to {n_{Mg{\text{ dư}}}} = \frac{1}{3} - 0,2 = \frac{2}{{15}}{\text{ mol}}\)

\( \to {m_{Mg}} = \frac{2}{{15}}.24 = 3,2\;{\text{gam}}\)

\( \to {n_{MgS{O_4}}} = {n_{{H_2}S{O_4}}} = 0,2{\text{ mol}}\)

\( \to {m_{MgS{O_4}}} = 0,2.(24 + 96) = 24{\text{ gam}}\) 

2)

Phản ứng xảy ra:

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

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

Ta có:

\({n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol = }}{{\text{n}}_{Ca}}\)

\( \to {m_{Ca}} = 0,1.40 = 4{\text{ gam}} \to {{\text{m}}_{CaO}} = 9,6 - 4 = 5,6{\text{ gam}}\)

\( \to {n_{CaO}} = \frac{{5,6}}{{40 + 16}} = 0,1{\text{ mol}}\)

\(\% {m_{Ca}} = \frac{4}{{9,6}} = 41,67\%  \to \% {m_{CaO}} = 58,33\% \)

\({n_{Ca{{(OH)}_2}}} = {n_{Ca}} + {n_{CaO}} = 0,1 + 0,1 = 0,2{\text{ mol}}\)

\( \to {m_{Ca{{(OH)}_2}}} = 0,2.(40 + 17.2) = 14,8{\text{ gam}}\)

3)

Phản ứng xảy ra:

\(F{\text{e}} + 2HCl\xrightarrow{{}}F{\text{e}}C{l_2} + {H_2}\)

Ta có:

\({n_{F{\text{e}}}} = \frac{{11,2}}{{56}} = 0,2{\text{ mol;}}{{\text{n}}_{HCl}} = 0,1.2 = 0,2{\text{ mol < 2}}{{\text{n}}_{F{\text{e}}}}\)

Vậy \(Fe\) dư

\( \to {n_{{H_2}}} = \frac{1}{2}{n_{HCl}} = {n_{F{\text{e}}C{l_2}}} = \frac{1}{2}.0,2 = 0,1{\text{ mol}}\)

\( \to {V_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)

\({n_{F{\text{e dư}}}} = 0,2 - 0,1 = 0,1{\text{ mol}}\)

\( \to {m_{F{\text{e dư}}}} = 0,1.56 = 5,6{\text{ gam}}\)

\({C_{M{\text{ Fe}}C{l_2}}} = \frac{{0,1}}{{0,1}} = 1M\)