Đáp án:

Giải thích các bước giải:

`n_(H_2)=(2,24)/(22,4)=0,1(mol)`

a,

PT

`Fe_3O_4+4H_2` $\xrightarrow{t^o}$ `3Fe+4H_2O` `(1)`

b,

Theo PT `(1)`

`n_(Fe_3O_4)=1/4 n_(H_2)=1/4 . 0,1=0,025(mol)`

`->m_(Fe_3O_4)=0,025.232=5,8(g)`

`n_(Fe)=3/4 n_(H_2)=3/4 . 0,1=0,075(mol)`

`->m_(Fe)=0,075.56=4,2(g)`

c,

PT

`Fe+2HCl->FeCl_2+H_2` `(2)`

Theo PT `(2)`

`n_(H_2)=n_(Fe)=0,075(mol)`

`->V_(H_2 \text{lí thuyết})=0,075.22,4=1,68(l)`

Mà `H=80%`

`->V_(H_2 \text{thực tế})=1,68.80%=1,344(l)`

Đáp án:

\(\begin{array}{l}
b)\\
{m_{F{e_3}{O_4}}} = 5,8g\\
{m_{Fe}} = 4,2g\\
c)\\
{V_{{H_2}}} = 1,344l
\end{array}\)

Giải thích các bước giải:

 \(\begin{array}{l}
a)F{e_3}{O_4} + 4{H_2} \to 3Fe + 4{H_2}O\\
b)\\
{n_{{H_2}}} = 0,1mol\\
 \to {n_{F{e_3}{O_4}}} = \dfrac{1}{4}{n_{{H_2}}} = 0,025mol\\
 \to {n_{Fe}} = \dfrac{3}{4}{n_{{H_2}}} = 0,075mol\\
 \to {m_{F{e_3}{O_4}}} = 5,8g\\
 \to {m_{Fe}} = 4,2g\\
c)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = {n_{Fe}} = 0,075mol\\
 \to {V_{{H_2}}} = 0,075 \times 22,4 \times 80\%  = 1,344l
\end{array}\)