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`-` Phương trình 1: `-` `3Fe + 2O_2 \overset{t^o}\to Fe_3O_4`

`-` $n_{Fe}$ `= ` $\dfrac{8,4}{56}$ `= 0,15` `(mol)` 

`-` Theo phương trình 1 `-` $n_{Fe_3O_4}$ `= 0,05 ` `(mol)` 

`->` `m = `$m_{Fe_3O_4}$ `= 0,05 × 232 = 11,6` `(g)`

`-` Theo bài, ta có phương trình 2:

`-` `Fe_2O_3 + 3H_2SO_4 -> Fe_2(SO_4)_3 + 3H_2O`

`-` $n_{Fe_3O_4}$ `=` $\dfrac{0,5}{232}$ `=` $2,1×10^{-3}$

`-` Theo phương trình 2 $n_{Fe_2(SO_4)_3}$ `= ` $2,1×10^{-3}$

`->` $m_{muối \ Fe_2(SO_4)_3}$ `=` $2,1 × 10^{-3}$ `× 400 = 0,84` `(g)`

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:

\({m_{F{e_3}{O_4}}} + {m_{{H_2}S{O_4}}} = {m_{muối}} + {m_{{H_2}O}}\)

\( \to 0,5 + \frac{1}{{116}}.98 = {m_{muối}} + \frac{1}{{116}}.18\)

\( \to {m_{muối}} = 1,18966{\text{ gam}}\)