5 sao và ctlhn nha mình thank

`1)`

`n_{Fe_2(SO_4)_3}={40}/{400}=0,1(mol)`

`n_{SO_2}={2,24}/{22,4}=0,1(mol)`

Quy `Fe_xO_y` về `Fe:x(mol);O:y(mol)`

Bảo toàn Fe: `x=2n_{Fe_2(SO_4)_3}=0,2(mol)`

`Fe^0->Fe^{+3}+3e`

`O^0+2e->O^{-2}`

`S^{+6}+2e->S^{+4}`

Bảo toàn electron: `3x=2y+2n_{SO_2}`

`->y={3.0,2-2.0,1}/2=0,2(mol)`

`->x/y={0,2}/{0,2}=1`

`->x=y=1`

`->` Oxit là `FeO`

`2)`

Thí nghiệm 1: `(n_{CuO};n_{Al_2O_3};n_{Fe_3O_4})=(x;y;z)(mol)`

`->80x+102y+232z=10(1)`

`Al_2O_3+2NaOH->2NaAlO_2+H_2O`

`->m_{CR\ spu}=80x+232z=7,96(2)`

`(1)-(2)->y=0,02`

Thí nghiệm 2, do dùng gấp đôi thí nghiệm 1 nên:

`(n_{CuO};n_{Al_2O_3};n_{Fe_3O_4})=(2x;2y;2z)(mol)`

`n_{H_2SO_4}=0,385.1=0,385(mol)`

`CuO+H_2SO_4->CuSO_4+H_2O`

`Al_2O_3+3H_2SO_4->Al_2(SO_4)_3+3H_2O`

`Fe_3O_4+4H_2SO_4->Fe_2(SO_4)_3+FeSO_4+4H_2O`

Theo PT: `n_{H_2SO_4}=2x+3.2y+4.2z=0,385`

`->2x+8z=0,385-6.0,02=0,265(3)`

`(2)(3)->x=0,0125(mol);z=0,03(mol)`

`\%m_{Al_2O_3}={0,02.102}/{10}.100\%=20,4\%`

`\%m_{CuO}={0,0125.80}/{10}.100\%=10\%`

`\%m_{Fe_3O_4}=100-10-20,4=69,6\%`

$\rm V.$

$\rm 1.$

$\rm 2Fe_xO_y+(6x-2y)H_2SO_4 \rightarrow xFe_2(SO_4)_3+(3x-2y)SO_2+(6x-2y)H_2O$

$\rm n_{SO_2}=$ $\rm\dfrac{2,24}{22,4}=0,1(mol)$ 

Theo $\rm PTHH:$ $\rm n_{Fe_2(SO_4)_3}=$ $\rm\dfrac{x}{3x-2y}n_{SO_2}$ 

$\rm\rightarrow$ $\rm\dfrac{40}{400}=$ $\rm\dfrac{0,1x}{3x-2y}$ 

Thỏa mãn: $\rm x=1, y=1$

$\rm\rightarrow Oxit:FeO$

$\rm 2.$

$\rm Al_2O_3+2NaOH \rightarrow 2NaAlO_2+H_2O$

$\rm\rightarrow m_{Rắn}=m_{CuO}+m_{Fe_3O_4}$

Bảo toàn $O:$ $\rm n_{O(X)}=n_{H_2SO_4}=0,385(mol)$

$20(g)X$ có $\rm 0,385.16=6,16(g)Oxi$

Vậy $\rm 10(g)X$ có $\rm\dfrac{10.6,16}{20}=3,08(g)Oxi$

Đặt $\rm n_{CuO}=a(mol), n_{Al_2O_3}=b(mol), n_{Fe_3O_4}=c(mol)$

$\rm\rightarrow 80a+102b+232c=10(1)$

Mặt khác khi cho hỗn hợp tác dụng với NaOH thu được 7,96 gam chất rắn.

$\rm\rightarrow 80a+232c=7,96(2)$

Bảo toàn số mol nguyên tố $O$ trong $10(g)X:$

$\rm a+3b+4c=$  $\rm\dfrac{3,08}{16}=0,1925(3)$

Từ $(1),(2),(3)$ $\rm\rightarrow a=0,0125, b=0,02, c=0,03$

$\rm\rightarrow \%CuO=$ $\rm\dfrac{0,0125.80}{10}.100\%=10\%$ 

$\rm\rightarrow \%Al_2O_3=$ $\rm\dfrac{0,02.102}{10}.100\%=20,4\%$ 

$\rm\rightarrow \%Fe_3O_4=100\%-10\%-20,4\%=69,6\%$