1)

Gọi số mol \(Na;Fe\) lần lượt là \(x;y\)

\( \to 23x + 56y = 7,9{\text{ gam}}\)

Phản ứng xảy ra:

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

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

Ta có:

\({n_{C{l_2}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol}}\)

\( \to {n_{C{l_2}}} = \frac{1}{2}{n_{Na}} + \frac{3}{2}{n_{Fe}} = 0,5x + 1,5y = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol}}\)

Giải được:

\(x=y=0,1\).

\( \to {m_{Na}} = 0,1.23 = 2,3{\text{ gam;}}{{\text{m}}_{Fe}} = 0,1.56 = 5,6{\text{ gam}}\)

\( \to \% {m_{Na}} = \frac{{2,3}}{{7,9}} = 29,11\%  \to \% {m_{Fe}} = 70,89\% \)

2)

Ta có:

\({n_{Fe}} = \frac{{16,8}}{{56}} = 0,3{\text{ mol;}}{{\text{n}}_{{O_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol}}\)

\(3Fe + 2{O_2}\xrightarrow{{{t^o}}}F{e_3}{O_4}\)

Ta có:

\({n_{Fe}} > \frac{3}{2}{n_{{O_2}}} \to {n_{F{e_3}{O_4}}} = \frac{1}{2}{n_{{O_2}}} = 0,05{\text{ mol}}\)

\( \to {n_{Fe{\text{ dư}}}} = 0,3 - \frac{3}{2}{n_{{O_2}}} = 0,15{\text{ mol}}\)

\(F{e_3}{O_4} + 4{H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + F{e_2}{(S{O_4})_3} + 4{H_2}O\)

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

\( \to {n_{FeS{O_4}}} = {n_{F{e_3}{O_4}}} + {n_{Fe}} = 0,15 + 0,05 = 0,2{\text{ mol}}\)

\({n_{F{e_2}{{(S{O_4})}_3}}} = {n_{F{e_3}{O_4}}} = 0,05{\text{ mol}}\)

\( \to {m_{muối}} = 0,2.(56 + 96) + 0,05.(56.2 + 96.3) = 50,4{\text{ gam}}\)

3)

Gọi số mol \(Fe;S\) lần lượt là \(x;y\).

\( \to 56x + 32y = 20\)

Phản ứng xảy ra:

\(3Fe + 2{O_2}\xrightarrow{{{t^o}}}F{e_3}{O_4}\)

\(S + {O_2}\xrightarrow{{{t^o}}}S{O_2}\)

Ta có:

\({n_{{O_2}}} = \frac{2}{3}{n_{Fe}} + {n_S} = \frac{2}{3}x + y = \frac{{6,72}}{{22,4}} = 0,3\)

Giải được: \(x=0,3;y=0,1\).

\( \to {m_{Fe}} = 0,3.56 = 16,8{\text{ gam;}}{{\text{m}}_S} = 0,1.32 = 3,2{\text{ gam}}\)

\( \to \% {m_{Fe}} = \frac{{16,8}}{{20}} = 84\%  \to \% {m_S} = 16\% \)