1)
Phản ứng xảy ra:
\(3Fe + 2{O_2}\xrightarrow{{{t^o}}}F{e_3}{O_4}\)
Ta có:
\({n_{Fe}} = \frac{{8,4}}{{56}} = 0,15{\text{ mol}}\)
\( \to {n_{F{e_3}{O_4}}} = \frac{1}{3}{n_{Fe}} = 0,05{\text{ mol}}\)
\( \to m = {m_{F{e_3}{O_4}}} = 0,05.(56.3 + 16.4) = 11,6{\text{ gam}}\)
2)
Phản ứng xảy ra:
\(F{e_3}{O_4} + 4{H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + F{e_2}{(S{O_4})_3} + 4{H_2}O\)
Ta có:
\({n_{F{e_3}{O_4}}} = \frac{{0,5}}{{232}}\)
\( \to {n_{{H_2}S{O_4}}} = {n_{{H_2}O}} = 4{n_{F{e_3}{O_4}}} = \frac{{0,5}}{{232}}.4 = \frac{1}{{116}}\)
BTKL:
\({m_{F{e_3}{O_4}}} + {m_{{H_2}S{O_4}}} = {m_{muối}} + {m_{{H_2}O}}\)
\( \to 0,5 + \frac{1}{{116}}.98 = {m_{muối}} + \frac{1}{{116}}.18\)
\( \to {m_{muối}} = 1,18966{\text{ gam}}\)
