1)

\({m_{dd}} = {m_{N{a_2}O}} + {m_{{H_2}O}} = 6,2 + 73,8 = 80{\text{ gam}}\)

\(N{a_2}O + {H_2}O\xrightarrow{{}}2NaOH\)

Ta có: 

\({n_{N{a_2}O}} = \frac{{6,2}}{{23.2 + 16}} = 0,1{\text{ mol}} \to {{\text{n}}_{NaOH}} = 2{n_{N{a_2}O}} = 0,2{\text{ mol}}\)

\({m_{NaOH}} = 0,2.40 = 8{\text{ gam}} \to {\text{C}}{{\text{% }}_{NaOH}} = \frac{8}{{80}} = 10\% \)

2)

Phản ứng xảy ra:

\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)

Ta có: 

\({n_{Fe}} = \frac{{11,2}}{{56}} = 0,2{\text{ mol; }}{{\text{n}}_{{H_2}S{O_4}}} = 0,2.3 = 0,6{\text{ mol > }}{{\text{n}}_{Fe}}\)

Vậy axit dư

\( \to {n_{FeS{O_4}}} = {n_{Fe}} = 0,2{\text{ mol; }}{{\text{n}}_{{H_2}S{O_4}{\text{ dư}}}} = 0,6 - 0,2 = 0,4{\text{ mol}}\)

\( \to {C_{M{\text{ FeS}}{{\text{O}}_4}}} = \frac{{0,2}}{{0,2}} = 1M;{C_{M{\text{ }}{{\text{H}}_2}S{O_4}{\text{ dư}}}} = \frac{{0,4}}{{0,2}} = 2M\)

3)

Gọi số mol Fe và Al lần lượt là x, y.

\( \to 56x + 27y = 11{\text{ gam}}\)

Phản ứng xảy ra:

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

\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)

\( \to {n_{{H_2}}} = \frac{3}{2}{n_{Al}} + {n_{Fe}} = 1,5y + x = \frac{{8,96}}{{22,4}} = 0,4{\text{ mol}}\)

Giải được: x=0,1; y=0,2

\( \to {m_{Fe}} = 0,1.56 = 5,6{\text{ gam}}\)

\( \to \% {m_{Fe}} = \frac{{5,6}}{{11}} = 50,9\%  \to \% {m_{Al}} = 49,1\% \)

\({n_{HCl}} = 2{n_{{H_2}}} = 0,8{\text{ mol}} \to {{\text{V}}_{HCl}} = \frac{{0,8}}{2} = 0,4{\text{ lít}}\)