PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

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

\(\Rightarrow160x+80y=40\left(1\right)\)

Ta có: \(n_{H_2}=\dfrac{14,56}{22,4}=0,65\left(mol\right)\)

Theo PT: \(n_{H_2}=3n_{Fe_2O_3}+n_{CuO}=3x+y\left(mol\right)\)

⇒ 3x + y = 0,65 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,15\left(mol\right)\\y=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe_2O_3}=\dfrac{0,15.160}{40}.100\%=60\%\\\%m_{CuO}=40\%\end{matrix}\right.\)

Bạn tham khảo nhé!

a)

Phần 1 : 

Gọi $n_{CuO} = a ; n_{Fe_2O_3} = b$

Ta có : 

$80a + 160b = 80 : 2 = 40(1)$
$CuO + 2HCl \to CuCl_2 + H_2O$
$Fe_2O_3 + 6HCl \to 2FeCl_3 + 3H_2O$

Suy ra:

$135a + 162,5.2b = 78,5(2)$
Từ (1)(2) suy ra a = 0,1 ; b = 0,2

$\%m_{CuO} = \dfrac{0,1.80}{40}.100\% = 20\%$
$\%m_{Fe_2O_3} = 80\%$

b)

Dung dịch muối gồm : 

$Fe^{3+} : 2b = 0,4(mol)$
$Cu^{2+} : a = 0,1(mol)$
$Cl^- : x(mol)$
$SO_4^{2-} : y(mol)$
Bảo toàn điện tích : 0,4.3 + 0,1.2 = x + 2y

$m_{muối} = 0,4.56 + 0,1.64 + 35,5x + 96y = 84,75$

Suy ra x = 0,9 ; y = 0,25

$C_{M_{HCl}} = \dfrac{0,9}{0,5} = 1,8M$
$C_{M_{H_2SO_4}} = \dfrac{0,25}{0,5} = 0,5M$

PTHH: \(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

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

Ta có: \(\%m_{ZnO}=20\%\) \(\Rightarrow\%m_{CuO}=80\%\)

Mặt khác: \(\left\{{}\begin{matrix}n_{Zn}=\dfrac{25\cdot20\%}{81}=\dfrac{5}{81}\left(mol\right)\\n_{CuO}=\dfrac{25\cdot80\%}{80}=0,25\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl}=\left(\dfrac{5}{81}+0,25\right)\cdot2=\dfrac{101}{162}\left(mol\right)\) 

\(\Rightarrow m_{ddHCl}=\dfrac{\dfrac{101}{162}\cdot36,5}{15\%}\approx151,71\left(g\right)\) \(\Rightarrow V_{ddHCl}=\dfrac{151,71}{1,1}\approx137,92\left(ml\right)\)             

a)%mZnO=20% => %mCuO=100% - 20%=80%

b) mZnO=20%.25=5(g)=> nZnO=5/81(mol)

mCuO=25-5=20(g) => nCuO=20/80=0,25(mol)

PTHH: ZnO +2 HCl -> ZnCl2 + H2O

5/81_______10/81___5/81(mol)

CuO +2 HCl -> CuCl2 + H2O

0,25__0,5______0,25(mol)

 => nHCl=10/81 + 0,5=101/162(mol)

=>mHCl= 101/162 . 36,5=7373/324(g)

=> mddHCl= 7373/324 : 15%= 151,708(g)

 => VddHCl= 151,708/1,1=137,916(ml)

Gọi \(\left\{{}\begin{matrix}n_{CuO}=x\\n_{Fe_2O_3}=y\end{matrix}\right.\)

\(n_{HCl}=0,5.2=1mol\)

\(CuO+2HCl\rightarrow CuCl_2+H_2\)

 x          2x                               ( mol )

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

   y             6y                                       ( mol )

Ta có:

\(\left\{{}\begin{matrix}80x+160y=32\\2x+6y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

\(\Rightarrow m_{CuO}=0,2.80=16g\)

\(\Rightarrow m_{Fe_2O_3}=0,1.160=16g\)

Gọi \(\left\{{}\begin{matrix}n_{CuO}=a\left(mol\right)\\n_{Fe_2O_3}=b\left(mol\right)\end{matrix}\right.\)

PTHH:

CuO + HCl ---> CuCl₂ + H₂O

a          2a

Fe₂O₃ + 6HCl ---> 2FeCl₃ + 3H₂O

b             6b

\(n_{HCl}=2.0,5=1\left(mol\right)\)

Hệ phương trình \(\left\{{}\begin{matrix}80a+160b=32\\2a+6b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,1\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{CuO}=0,2.80=16\left(g\right)\\m_{Fe_2O_3}=0,1.160=16\left(g\right)\end{matrix}\right.\)

\(n_{HCl}=0,5.1=0,5\left(mol\right)\)

a)

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

x------->2x

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

y--------->6y

Có hệ: \(\left\{{}\begin{matrix}2x+6y=0,5\\80x+160y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

\(m_{CuO}=0,1.80=8\left(g\right)\\ m_{Fe_2O_3}=0,05.160=8\left(g\right)\)

b

\(\%m_{CuO}=\dfrac{0,1.80.100\%}{16}=50\%\\ \%m_{Fe_2O_3}=\dfrac{0,05.160.100\%}{16}=50\%\)

a,\(n_{HCl}=0,5.1=0,5\left(mol\right)\)

PTHH: CuO + 2HCl → CuCl₂ + H₂O

Mol:       x          2x

PTHH: Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O

Mol:        y            6y

Ta có: \(\left\{{}\begin{matrix}80x+160y=16\\2x+6y=0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

PTHH: CuO + 2HCl → CuCl₂ + H₂O

Mol:      0,1       0,2

PTHH: Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O

Mol:      0,05         0,3

\(\Rightarrow m_{CuO}=0,1.80=8\left(g\right);m_{Fe_2O_3}=16-8=8\left(g\right)\)

b,\(\%m_{CuO}=\dfrac{8.100\%}{16}=50\%;\%m_{Fe_2O_3}=100-50=50\%\)

Gọi $n_{CaO} = a(mol) ; n_{Fe_2O_3} = b(mol) \Rightarrow 56a + 160b = 24,4(1)$

$CaO + 2HCl \to CaCl_2 + H_2O$
$Fe_2O_3 + 6HCl \to 2FeCl_3 + 3H_2O$
$n_{HCl} = 2a + 6b = 0,45. 2= 0,9(2)$

Từ (1)(2) suy ra a = 0,15;  b = 0,1

$m_{CaO} = 0,15.56 = 8,4(gam)$

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

Ta có: 80n_CuO + 160n_Fe2O3 = 24 (1)

\(n_{H_2SO_4}=0,2.2=0,4\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{CuO}+3n_{Fe_2O_3}=0,4\left(2\right)\)

Từ (1) và (2) ⇒ n_CuO = n_Fe2O3 = 0,1 (mol)

⇒ m_CuO = 0,1.80 = 8 (g)

m_Fe2O3 = 0,1.160 = 16 (g)

\(n_{CuO}=a\left(mol\right);n_{Fe_2O_3}=b\left(mol\right)\left(a,b>0\right)\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ \Rightarrow\left\{{}\begin{matrix}80a+160b=24\\a+3b=2.0,2=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \%m_{CuO}=\dfrac{0,1.80}{24}.100\%\approx33,333\%;\%m_{Fe_2O_3}=\dfrac{160.0,1}{24}.100\%\approx66,667\%\)

đổi `200ml=0,2l`

`=>n_(H_2 SO_4)=C_M *V=2*0,2=0,4(mol)`

gọi: \(\left\{{}\begin{matrix}n_{CuO}=a\left(mol\right)\\n_{Fe_2O_3}=b\left(mol\right)\end{matrix}\right.\)

\(PTHH:CuO+H_2SO_4->CuSO_4+H_2O\)

tỉ lệ          1        :      1         :         1       :   1

n(mol)      a----------->a---------->a------------>a

\(PTHH:Fe_2O_3+3H_2SO_4->Fe_2\left(SO_4\right)_3+3H_2O\)

tỉ lệ             1      ;      3          :           1           :     3

n(mol)        b--------->3b------------>b-------------->3b

Ta có hệ phương trình sau

\(\left\{{}\begin{matrix}80a+160b=24\\a+3b=0,4\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ =>\left\{{}\begin{matrix}n_{CuO}=0,1\left(mol\right)\\n_{Fe_2O_3}=0,1\left(mol\right)\end{matrix}\right.\\ =>\left\{{}\begin{matrix}m_{CuO}=n\cdot M=0,1\cdot80=8\left(g\right)\\m_{Fe_2O_3}=n\cdot M=0,1\cdot160=16\left(g\right)\end{matrix}\right.\)