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

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

Giả sử: \(\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{Fe_2O_3}=y\left(mol\right)\end{matrix}\right.\)

Ta có: \(n_{HCl}=0,1.5=0,5\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{CuO}+6n_{Fe_2O_3}=2x+6y\left(mol\right)\) ⇒ 2x + 6y = 0,5 (1)

\(\left\{{}\begin{matrix}n_{CuCl_2}=n_{CuO}=x\left(mol\right)\\n_{FeCl_3}=2n_{Fe_2O_3}=2y\left(mol\right)\end{matrix}\right.\)

Mà: n_CuCl2 : n_FeCl3 = 1:1

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

⇒ a = m_CuO + m_Fe2O3 = 0,1.80 + 0,05.160 = 16 (g)

Bạn tham khảo nhé!

\(n_{CuCl_2}=n_{FeCl_3}=a\left(mol\right)\)

\(n_{HCl}=0.1\cdot5=0.5\left(mol\right)\)

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

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

\(n_{HCl}=2a+3a=0.5\left(mol\right)\)

\(\Rightarrow a=0.1\)

\(m_{hh}=0.1\cdot80+0.05\cdot160=16\left(g\right)\)

 hỗn hợp trên là oxit...

a, Ta có

CuO + 2HCl \(\rightarrow\) CuCl₂ + H₂O

x \(\rightarrow\) 2x \(\rightarrow\) x \(\rightarrow\) x

Fe₂O₃ + 6HCl ​\(\rightarrow\) 2FeCl₃ + 3H₂O

y \(\rightarrow\) 6y \(\rightarrow\) 2y \(\rightarrow\) 3y

Theo 2 phương trình trên ta có

n_CuCl2 / n_FeCl3 = 1/1 => x / 2y = 1/1

=> x = 2y => x - 2y = 0

=> \(\left\{{}\begin{matrix}80x+160y=8\\\text{x - 2y = 0}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=0,05\\y=0,025\end{matrix}\right.\)

=> M_HCl = ( 2x + 6y ) . 36,5 = 9,125 ( gam )

b, 200 ml = 0,2 l

=> C_{M HCl} = n : V = ( 2x + 6y ) : 0,2 = 1,25 M

không chép trên mạng xuống nhe mấy bạn 

Fe3O4 + 8HCl --> FeCl2 + 2FeCl3 + 4H2O
_a_______8a______a_____2a______4a_
Fe2O3 + 6HCl --> 2FeCl3 + 3H2O 
_b______6b_______2b_____3b_
FeO + 2HCl --> FeCl2 + H2O 
_c____2c_______c______c_
Fe + 2HCl --> FeCl2 + H2
_d___2d________d________d_

H2 + CuO --> Cu + H2O
_d____d_____d_____d_
nCuO = 3.2 / 80 = 0.04
=> d = 0.04

mHCl = 360 * 18.25 / 100 = 65.7 (g)
nHCl = 65.7 / 36.5 = 1.8 (mol)
=> 8a + 6b + 2c + 2d = 1.8
=> 8a + 6b + 2c + 0.08 = 1.8
=> 8a + 6b + 2c = 1.72
=> 4a + 3b + c = 0.86

a)
theo đlbtkl ta có
mHCl + mhh = mmuối + mH2O + mH2
65.7 + 57.6 = mmuối + 0.86 * 18 + 0.04 * 2
=> mmuối = 107.74 (g)
b) Fe3O4 + 8HCl --> FeCl2 + 2FeCl3 + 4H2O
_a_______8a______a_____2a______4a_
Fe2O3 + 6HCl --> 2FeCl3 + 3H2O 
_b______6b_______2b_____3b_
FeO + 2HCl --> FeCl2 + H2O 
_b____2b_______b______b_
Fe + 2HCl --> FeCl2 + H2
0.2__0.4_______0.2____0.2
nH2 = nFe = 0.2
232a + 232b =57.6-0.2X56
8a + 8b = 1.8-0.2X2
vô nghiệm vì nFe2O3 =nFeO tuong đương 2ẩn cung M