C9 thui ạ...........

`a)`

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

`CuO+H_2`  $\xrightarrow{t^o}$  `Cu+H_2O`

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

`MgO+2HCl->MgCl_2+H_2O`

`CuO+2HCl->CuCl_2+H_2O`

`b)`

Thí nghiệm 1: `(n_{Fe_3O_4};n_{MgO};n_{CuO})=(x;y;z)(mol)`

`->232x+40y+80z=25,6(1)`

Theo PT: `n_{Fe}=3x(mol);n_{Cu}=z(mol)`

`->56.3x+40y+64z=20,8(2)`

Thí nghiệm 2: `(n_{Fe_3O_4};n_{MgO};n_{CuO})=(kx;ky;kz)(mol)`

`->kx+ky+kz=0,15(3)`

`225ml=0,225l`

`->n_{HCl}=0,225.2=0,45(mol)`

Theo PT: `8kx+2ky+2kz=n_{HCl}=0,45(4)`

Lấy `{(3)}/{(4)}->{x+y+z}/{8x+2y+2z}={0,15}/{0,45}=1/3`

`->5x-y-z=0(5)`

`(1)(2)(5)->x=0,05(mol);y=0,15(mol);z=0,1(mol)`

`\%m_{Fe_3O_4}={0,05.232}/{25,6}.100\%=45,3125\%`

`\%m_{MgO}={0,15.40}/{25,6}.100\%=23,4375\%`

`\%m_{CuO}=100-23,4735-45,3125=31,25\%`

Tham khảo :

$a,$

$Fe_3O_4+4H_2 \xrightarrow{t^o} 3Fe+4H_2O$

$CuO+H_2 \xrightarrow{t^o} Cu+H_2O$

$Fe_3O_4+8HCl \to 2FeCl_3+FeCl_2+4H_2O$$(1)$

$MgO+2HCl \to MgCl_2+H_2O$$(2)$

$CuO+2HCl \to CuCl_2+H_2O$$(3)$

$b,$

$n_{HCl}=0,45(mol)$

`-` Bảo toàn nguyên tố $H:$

$n_{H_2}=n_{H_2O}$

`-` Bảo toàn khối lượng:

$m_{X}+m_{H_2}=m_{rắn}+m_{H_2O}$

`=>`$n_{H_2}=\dfrac{25,6-20,8}{18-2}=0,3(mol)$

$n_{Fe_3O_4}=a(mol);n_{MgO}=b(mol);n_{CuO}=c(mol)$

`->`$232a+40b+80c=25,6$$(*)$

`-` Bảo toàn nguyên tố $Fe:$

$n_{Fe}=3n_{Fe_3O_4}=3a(mol)$

`-` Bảo toàn nguyên tố $Cu:$

$n_{Cu}=n_{CuO}=c(mol)$

`->`$56 \cdot 3a+40b+64c=20,8$$(**)$

`-` Từ phương trình phản ứng $(1),(2),(3)$

`->`$\dfrac{a+b+c}{8a+2b+2c}=\dfrac{0,15}{0,45}$

`->`$0,75a-0,15b-0,15c=0$$(***)$

`-` Từ $(*),(**),(***)$ suy ra $a=0,05;b=0,15;c=0,1$

`=>`$\%m_{Fe_3O_4}=45,3125\%;\%m_{MgO}=23,4375\%;\%m_{CuO}=31,25\%$