`a.` `PTHH``:`

          `Fe_2``O_3``+``3H_2` $\xrightarrow[]{t^o}$ `2Fe``+``3H_2O`

          `Fe``+``2HCl``→``FeCl_2``+``H_2`

`b.` $n_{H_2(đktc)}$`=``\frac{V}{22,4}``=``\frac{3,36}{22,4}``=``0,15` `(mol)`

`PTHH`            `Fe``+``2HCl``→``FeCl_2``+``H_2`

                       `0,15`                                        `0,15`       `mol`

`→``n_{Fe}``=``n_{H_2}``=``0,15` `(mol)`

`→``m_{Fe}``=``n``.``M``=``0,15``.``56``=``8,4` `(g)`

`PTHH`            `Fe_2``O_3``+``3H_2` $\xrightarrow[]{t^o}$ `2Fe``+``3H_2O`

                            `0,075`         `0,225`      `0,15`                   `mol`

`→``n_{Fe_2O_3}``=``n_{Fe}``=``\frac{0,15.1}{2}``=``0,075` `(mol)`

     `m_{Fe_2O_3}``=``n``.``M``=``0,075``.``160``=``12` `(g)`

`→``n_{H_2}``=``n_{Fe}``=``\frac{0,15.3}{2}``=``0,225` `(mol)`

     $V_{H_2(đktc)}$`=``n``.``22,4``=``0,225``.``22,4``=``5,04` `(l)`

Giải thích các bước giải+Đáp án:

a,  $n_{H2}$ =$\frac{3,36}{22,4}$ =0,15(mol)

Ta có phương trình như sau:

$Fe_{2}$ $O_{3}$ +$3H_{2}$ →$2Fe_{}$ +$3H_{2}O$ (1)

$Fe_{}$ +$2HCl_{}$ →$FeCl_{2}$ +$H_{2}$   (2)

b, 

Theo(1) có:

$n_{Fe2O3}$ =$\frac{1}{3}$ .$n_{H2}$=$\frac{1}{3}$ .0,15=0,05(mol)

⇒$m_{Fe2O3}$ = 0,1.160=16(g)

Ta có: $n_{H2}$ =$\frac{3,36}{22,4}$ =0,15(mol)

⇒$V_{H2}$ = 0,15.22,4=3,36(lít)

$n_{Fe}$= $\frac{2}{3}$ .$n_{H2}$=$\frac{2}{3}$ .0,15=0,1(mol)

⇒$m_{Fe}$ =0,1.56=5,6(g)

Tóm tắt lại ta có:

m=16g

v=3,36 lít

a=5,6g