Gọi: \(\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{Fe_2O_3}=y\left(mol\right)\\n_{MgO}=z\left(mol\right)\end{matrix}\right.\) ⇒ 80x + 160y + 40z = 12 (1)

- Cho X pư với dd HCl.

PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

Theo PT: \(n_{HCl}=2n_{CuO}+6n_{Fe_2O_3}+2n_{MgO}=2x+6y+2z=0,45\left(2\right)\)

- Cho CO qua hh nung nóng.

Có: \(kx+ky+kz=0,175\)

PT: \(CuO+CO\underrightarrow{t^o}Cu+CO_2\)

\(Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\)

Theo PT: \(\left\{{}\begin{matrix}n_{Cu}=n_{CuO}=kx\left(mol\right)\\n_{Fe}=2n_{Fe_2O_3}=2ky\left(mol\right)\end{matrix}\right.\)

⇒ 64kx + 56.2ky + 40kz = 10

Ta có: \(\dfrac{kx+ky+kz}{64kx+56.2ky+40kz}=\dfrac{0,175}{10}\) \(\Rightarrow\dfrac{x+y+z}{64x+112y+40z}=\dfrac{7}{400}\)

⇒ 6x + 48y - 15z = 0 (3)

Từ (1), (2) và (3) \(\Rightarrow\left\{{}\begin{matrix}x=0,05\left(mol\right)\\y=0,025\left(mol\right)\\z=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=0,05.80=4\left(g\right)\\m_{Fe_2O_3}=0,025.160=4\left(g\right)\\m_{MgO}=0,1.40=4\left(g\right)\end{matrix}\right.\)

Đáp án : A

$2Mg + O_2 \xrightarrow{t^o} 2MgO$
$2Cu + O_2 \xrightarrow{t^o} 2CuO$
$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$

$MgO + 2HCl \to MgCl_2 + H_2O$
$CuO + 2HCl \to CuCl_2 + H_2O$
$Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O$

Gọi $n_{MgO} = a(mol) ; n_{CuO} = b(mol) ; n_{Al_2O_3} = c(mol)$

Bảo toàn khối lượng : $m_{O_2} = 23,2 - 16,8 = 6,4(gam)$
$n_{O_2} = 0,2(mol)$

$\Rightarrow 0,5a + 0,5b + 1,5c = 0,2(1)$

Theo PTHH : 

$n_{HCl} =2 n_{MgO} + 2n_{CuO} + 6n_{Al_2O_3} = 0,8(theo (1))$

Suy ra : $V_{dd\ HCl} = \dfrac{0,8}{2} = 0,4(lít)$

Đáp án  A

Khi cho X tác dụng với HCl: n_HCl= 0,225.2= 0,45 mol

CuO + 2HCl→ CuCl₂+ H₂O

Fe₂O₃+ 6HCl→ 2FeCl₃+ 3H₂O

MgO + 2HCl→ MgCl₂+ H₂O

Đặt n_CuO = x mol; n F e 2 O 3 = y mol; n_MgO = z mol

→ 80x+ 160y +40z= 12 gam (1)

n_HCl= 2x+6y+2z= 0,45 mol (2)

C O     + C u O → t 0 C u                 + C O 2                           x                             x   m o l F e 2 O 3   +       3 C O → t 0 2 F e           + 3 C O 2 y                                                                   2 y   m o l

Chất rắn Y chứa x mol Cu; 2y mol Fe và z mol MgO

→ 64x + 56.2y + 40z= 10 gam (3)

Từ các PT(1,2,3) ta có x= 0,05; y=0,025; z=0,1

→% m F e 2 O 3 =33,33%

1) \(\left\{{}\begin{matrix}n_{CO}+n_{CO_2}=\dfrac{3,136}{22,4}=0,14\left(mol\right)\\\dfrac{28.n_{CO}+44.n_{CO_2}}{n_{CO}+n_{CO_2}}=16.2=32\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}n_{CO}=0,105\left(mol\right)\\n_{CO_2}=0,035\left(mol\right)\end{matrix}\right.\)

n_O = n_CO2 = 0,035 (mol)

=> a = 2,92 + 0,035.16 = 3,48(g)

\(n_{H_2SO_4}=\dfrac{100.5,39\%}{98}=0,055\left(mol\right)\)

n_H2O = n_O = 0,035 (mol)

Bảo toàn H: \(n_{H_2}=\dfrac{0,055.2-0,035.2}{2}=0,02\left(mol\right)\)

=> \(V=0,02.22,4=0,448\left(l\right)\)

2) Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Fe_3O_4}=b\left(mol\right)\\n_{CuO}=c\left(mol\right)\end{matrix}\right.\)

=> 56a + 232b + 80c = 3,48 (1)

Bảo toàn Fe: n_Fe = a + 3b (mol)

Bảo toàn Cu: n_Cu = c (mol)

PTHH: Fe + H₂SO₄ --> FeSO₄ + H₂

         0,02<-0,02<------0,02<---0,02

            Fe₃O₄ + 4H₂SO₄ --> Fe₂(SO₄)₃ + FeSO₄ + 4H₂O

                 b--->4b------------>b-------------->b

             CuO + H₂SO₄ --> CuSO₄ + H₂O

                c---->c------------>c

=> a = 0,02

=> 0,02 + 4b + c = 0,055 => 4b + c = 0,035 

(1) => 232b + 80c = 2,36

=> b = 0,005 (mol); c = 0,015 (mol)

B chứa \(\left\{{}\begin{matrix}FeSO_4:0,025\left(mol\right)\\Fe_2\left(SO_4\right)_3:0,005\left(mol\right)\\CuSO_4:0,015\left(mol\right)\end{matrix}\right.\)

m_{dd sau pư} = 3,48 + 100 - 0,02.2 = 103,44 (g)

=> \(\left\{{}\begin{matrix}C\%_{FeSO_4}=\dfrac{0,025.152}{103,44}.100\%=3,674\%\\C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,005.400}{103,44}.100\%=1,933\%\\C\%_{CuSO_4}=\dfrac{0,015.160}{103,44}.100\%=2,32\%\end{matrix}\right.\)

3)

Rắn khan chứa \(\left\{{}\begin{matrix}BaSO_4\\Fe\left(OH\right)_3\\Cu\left(OH\right)_2\end{matrix}\right.\)

Có: \(n_{BaSO_4}=n_{SO_4}=0,055\left(mol\right)\)

Bảo toàn Fe: \(n_{Fe\left(OH\right)_3}=n_{FeSO_4}+2.n_{Fe_2\left(SO_4\right)_3}=0,035\left(mol\right)\)

Bảo toàn Cu: \(n_{Cu\left(OH\right)_2}=0,015\left(mol\right)\)

=> b = 0,055.233 + 0,035.107 + 0,015.98 = 18,03 (g)