1)
Phản ứng xảy ra:
\(N{a_2}O + {H_2}O\xrightarrow{{}}2NaOH\)
Ta có:
\({n_{N{a_2}O}} = \frac{{6,2}}{{23.2 + 16}} = 0,1{\text{ mol}}\)
\( \to {n_{NaOH}} = 2{n_{N{a_2}O}} = 0,1.2 = 0,2{\text{ mol}}\)
\( \to {C_{M{\text{ NaOH}}}} = \frac{{0,2}}{{0,5}} = 0,4M\)
2)
Phản ứng xảy ra:
\(F{\text{e}} + 2HCl\xrightarrow{{}}F{\text{e}}C{l_2} + {H_2}\)
Ta có:
\({n_{HCl}} = 0,2.0,2 = 0,04{\text{ mol}}\)
Theo phản ứng:
\({n_{F{\text{e}}C{l_2}}} = {n_{F{\text{e}}}} = {n_{{H_2}}} = \frac{1}{2}{n_{HCl}} = 0,02{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,02.22,4 = 0,448{\text{ lít}}\)
\({m_{F{\text{e}}}} = 0,02.56 = 1,12{\text{ gam}}\)