Khu m gam sat 3 oxit bang V lit khi H2 thu dc a gam fe. Cho toan bo luong fe nay tac dung voi dd hlc du , thu dc 3.36 lit khi h2 (dktc) a. viet pthh b. tinh m, v, a

`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)`