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: \(x=0,3;y=0,1\).
\( \to {m_{Fe}} = 0,3.56 = 16,8{\text{ gam;}}{{\text{m}}_S} = 0,1.32 = 3,2{\text{ gam}}\)
\( \to \% {m_{Fe}} = \frac{{16,8}}{{20}} = 84\% \to \% {m_S} = 16\% \)
