1)

Phản ứng xảy ra:

\(NaC{l_r} + {H_2}S{O_4}\xrightarrow{{}}NaH{\text{S}}{O_4} + HCl\)

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

\(F{\text{e}}C{l_2} + 2NaOH\xrightarrow{{}}F{\text{e}}{(OH)_2} + 2NaCl\)

\(2NaCl\xrightarrow{{đpnc}}2Na + C{l_2}\)

\(C{l_2} + {H_2}O\overset {} \leftrightarrows HCl + HClO\)

2)

Phản ứng xảy ra:

\(Mn{O_2} + 4HCl\xrightarrow{{}}MnC{l_2} + C{l_2} + 2{H_2}O\)

\(2NaOH + C{l_2}\xrightarrow{{}}NaCl + NaClO + {H_2}O\)

Ta có:

\({n_{Mn{O_2}}} = \frac{{34,8}}{{55 + 16.2}} = 0,4{\text{ mol}}=n_{Cl_2}\)

\({n_{NaOH}} = 0,25.4 = 1{\text{ mol > 2}}{{\text{n}}_{C{l_2}}}\)

Vậy \(NaOH\) dư

\( \to {n_{NaCl}} = {n_{NaClO}} = {n_{C{l_2}}} = 0,4{\text{ mol;}}{{\text{n}}_{NaOH{\text{ dư}}}} = 1 - 0,4.2 = 0,2{\text{ mol}}\)

\( \to {C_{M{\text{ NaCl}}}} = {C_{M{\text{ NaClO}}}} = \frac{{0,4}}{{0,25}} = 1,6M;{C_{M{\text{ NaOH}}}} = \frac{{0,2}}{{0,25}} = 0,8M\)