1)  \(2Na + {I_2}\xrightarrow{{}}2NaI\)

 Ta có:  \({n_{NaI}} = \frac{{30}}{{23 + 127}} = 0,2{\text{ mol}} \to {{\text{n}}_{Na}} = {n_{NaI}} = 0,2{\text{ mol}} \to {{\text{m}}_{Na}} = 0,2.23 = 4,6{\text{ gam}}\)

\({n_{{I_2}}} = \frac{1}{2}{n_{NaI}} = 0,1{\text{ mol}} \to {{\text{m}}_{{I_2}}} = 0,1.127.2 = 25,4{\text{ gam}}\)

2)  \(2Fe + 3C{l_2}\xrightarrow{{}}2FeC{l_3}\)

\({n_{Fe}} = \frac{{22,4}}{{56}} = 0,4{\text{ mol = }}{{\text{n}}_{FeC{l_3}}} \to {m_{FeC{l_3}}} = 0,4.(56 + 35,5.3) = 65{\text{ gam}}\)

\({n_{C{l_2}}} = \frac{3}{2}{n_{Fe}} = 0,6{\text{ mol}} \to {{\text{V}}_{C{l_2}}} = 0,6.22,4 = 13,34{\text{ lít}}\)

3) \(2Na + {X_2}\xrightarrow{{}}2NaX\)

\({n_{Na}} = {n_{NaX}} \to \frac{{2,3}}{{23}} = \frac{{5,85}}{{23 + X}} \to X = 35,5 \to X:{\text{ Cl}}\)

4) \(Mg + {X_2} \to Mg{X_2}\)

\({n_{Mg}} = {n_{{X_2}}} = \frac{6}{{24}} = 0,25{\text{ mol}} \to {{\text{M}}_{{X_2}}} = \frac{{40}}{{0,25}} = 160 \to {M_X} = 80 \to X:Br\)