$\rm m_{không~tan}=m_{Cu(dư)}=8,32(g)$

Đặt $\rm a(mol)=n_{Fe_3O_4}$ 

$\rm Fe_3O_4 +8HCl \xrightarrow{} FeCl_2 +2FeCl_3 +4H_2O$ 

$\rm Cu +2FeCl_3 \xrightarrow{} CuCl_2 +2FeCl_2$ 

Theo phương trình hóa học 

$\rm n_{FeCl_2}=3a(mol)$

$\rm n_{CuCl_2}=\dfrac{1}{2}n_{FeCl_3}=a(mol)$ 

Ta có $\rm m_{rắn}=m_{FeCl_2}+m_{CuCl_2}$ 

`<=> `$\rm 61,92=127.3.a+136a$

`<=>`$\rm a=0,12(mol)$ 

$\rm m=m_{Fe_3O_4} +m_{Cu(phản~ứng)} +m_{Cu(dư)}\\~~~=0,12.232+0,12.64+8,32=43,84(g)$

Đặt `n_{Fe_3O_4} = x(mol)`

`Fe_3O_4 + 8HCl -> 2FeCl_3 + FeCl_2 + 4H_2O(1)`

`x -> 8x -> 2x -> x -> 4x(mol)`

` 2FeCl_3 + Cu -> 2FeCl_2 + CuCl_2(2)`

`2x -> x -> 2x -> x(mol)`

Rắn khan:

`{(FeCl_2: 3x(mol)),(CuCl_2: x(mol)):}`

`=> 127 . 3x + 135x = 61,92`

`=> x = 0,12`

`=> m = m_{Fe_3O_4} + m_{Cu(p//ứ)} + m_{Cu(dư)}`

`=> m = 0,12 . 232 + 0,12 . 64 + 8,32 = 43,84g`