Đáp án:

$\rm a)\ V_{O_2}=1,68(l)$

$\rm b)\ m_{FeCl_2}=19,05(g)$

$\rm c)\ C$`%` $\rm _{dd\ HCl}$`=3,65%`

$\rm d)\ C$`%` $\rm _{dd\ FeCl_2}$`~~6,13%`

Giải thích các bước giải:

$\rm a)$

            $\rm 2Fe+O_2$ $\xrightarrow{\rm t^o}$ $\rm 2FeO$

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

$\rm n_{O_2}=\dfrac{1}{2}n_{Fe}=0,075(mol)$

$\rm V_{O_2}=0,075.22,4=1,68(l)$

$\rm b)$

             $\rm FeO+2HCl\to FeCl_2+H_2O$

$\rm n_{FeCl_2}=n_{FeO}=n_{Fe}=0,15(mol)$

$\rm m_{FeCl_2}=0,15.127=19,05(g)$

$\rm c)$

$\rm n_{HCl}=2n_{FeO}=0,3(mol)$

$\rm C$`%` $\rm _{dd\ HCl}$`={0,3.36,5}/{300}.100%=3,65%`

$\rm d)$

$\rm C$`%` $\rm _{dd\ FeCl_2}$`={19,05}/{(0,15.72)+300}.100%~~6,13%`

Đáp án `+` Giải thích các bước giải:

`a)`

`n_(Fe) = (8,4)/(56)=0,15` `(mol)`

`2Fe +  O_2` $\xrightarrow{t^o}$ `2FeO`

`n_(O_2) =1/2n_(Fe)=0,075` `(mol)`

`V_(O_2) = 0,075xx22,4 = 1,68` `(l)`

`b)`

`n_(FeO) = n_(Fe) = 0,15` `(mol)`

`FeO + 2HCl -> FeCl_2 + H_2O`

`n_(FeCl_2) = n_(FeO) = 0,15` `(mol)`

`m_(FeCl_2) = 0,15xx127 = 19,05` `(g)`

`c)`

`n_(HCl)= 2n_(FeO) = 0,3` `(mol)`

`C%_(HCl) = (0,3xx36,5xx100%)/(300) = 3,65%`

`d)` 

`m_(\text(dd sau pư)) = 0,15xx72 + 300 = 310,8` `(g)`

`C%_(FeCl_2) = (19,05xx100%)/(310,8) = 6,13%`