bài 1 , đổi : 800ml=0,8l

a, CuO + 2HCl---> CuCl2 + H2O(1)

Al2O3+ 6HCl---> 2AlCl3 + 3H2O(2)

b, Gọi nCuO=a mol , nAl2O3=b mol(a, b>0)

Ta có mCuO+ mAl2O3=80a+ 102b=10,12g

Theo PTHH ta có nHCl=nHCl(1)+ nHCl(2)=2nCuO+ 6nAl2O3=2a+6b

=0,575.0,8=0,46mol

Ta có 80a+ 102b=10,12

và 2a+ 6b=0,46mol

=> a=0,05mol; b=0,06mol

=>%mCuO=0,06.80.100/10,12=47,43%

%mAl2O3=100%-47,43%=52,57%

c, Theo PTHH ta có nCuCl2=nCuO=0,05mol

nAlCl3=2nAl2O3=0,06.2=0,12mol

=> CmCuCl2=0,05/0,8=0,0625M

CmAlCl3=0,12/0,8=0,15M

Câu 1:

\(a\text{) }CuO+2HCl\rightarrow CuCl_2+H_2O\left(1\right)\\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }x\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }2x\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }x\\ Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\left(2\right)\\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }y\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }6y\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }2y\)

b) \(n_{HCl}=C_M\cdot V=0,8\cdot0,575=0,46\left(mol\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\), ta có hệ pt: \(\left\{{}\begin{matrix}2x+6y=0,46\\80x+102y=10,12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,06\end{matrix}\right.\)

\(\Rightarrow m_{CuO}=n\cdot M=0,05\cdot80=4\left(g\right)\\ m_{Al_2O_3}=n\cdot M=0,06\cdot102=6,12\left(g\right)\)

\(\Rightarrow\%CuO=\dfrac{4}{10,12}\cdot100=39,53\%\\ \%Al_2O_3=\dfrac{6,12}{10,12}\cdot100=60,47\%\)

\(\text{c) }C_{M\left(CuCl_2\right)}=\dfrac{n}{V}=\dfrac{0,05}{0,8}=0,0625\left(M\right)\\ C_{M\left(AlCl_3\right)}=\dfrac{n}{V}=\dfrac{0,06\cdot2}{0,8}=0,15\left(M\right)\)

Câu 2:

\(n_{CuO}=\dfrac{m}{M}=\dfrac{6,4}{80}=0,08\left(mol\right)\\ n_{HCl}=C_M\cdot V=0,36\cdot0,6=0,216\left(mol\right)\)

\(\Rightarrow C_{M\left(HCl_{dư}\right)}=\dfrac{n}{V}=\dfrac{0,056}{0,36}=0,156\left(M\right)\\ C_{M\left(CuCl_2\right)}=\dfrac{n}{V}=\dfrac{0,08}{0,36}=0,223\left(M\right)\)

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

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

b, Gọi: \(\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{Fe_2O_3}=y\left(mol\right)\end{matrix}\right.\) ⇒ 80x + 160y = 32 (1)

Theo PT: \(\left\{{}\begin{matrix}n_{CuCl_2}=n_{Cu}=x\left(mol\right)\\n_{FeCl_3}=2n_{Fe_2O_3}=2y\left(mol\right)\end{matrix}\right.\) ⇒ 135x + 325y = 59,5 (2)

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

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

c, Theo PT: \(n_{HCl}=2n_{CuO}+6n_{Fe_2O_3}=1\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{1}{0,5}=2\left(l\right)\)

CuO +2HCl= CuCl2 +H2O 
ZnO+2HCl= ZnCl2 +H2O 
gọi x,y là mol của CuO, ZnO 
80x + 81y = 12.1 
2x+2y = 0.3 
=> x=0.05 , y=0.1 => mCuO= 4 %CuO=4/12.1 m ZnO=8.1 =>%ZnO=8.1/12.1 
nH2SO4=1/2nHCl=0.3/2 =0.15 
mH2SO4=0.15x98=14.7g => mddH2SO4=14.7/20%=73.5g

nHCl=0,4.0,1=0,04(mol)

Đặt : nCaO=x(mol); nCuO=y(mol)

\(\Rightarrow\) 56x+80y=1,36(1)

PT: CaO+2HCl \(\rightarrow\) CaCl2+H2O

x 2x x (mol)

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

y 2y y (mol)

nHCl=0,04 mol \(\Rightarrow\) 2x+2y=0,04(2)

Từ (1) và (2) giải hệ pt tìm đk : x=y=0,01

%m CaO=41,2% ; %mCuO=58,8%

mCaCl2=0,01.111=1,11g

mCuCl2=0,01.135=1,35 g

nHCL=CM.V=0.1.0.4=0.04(mol)

a. PTHH: CaO + 2HCL suy ra CaCL2 +H20 (1)

x 2x x (mol)

CuO + 2HCL suy ra CuCL2 +H2O (2)

y 2y y (mol)

Gọi nCaO=x (mol) theo pt(1): nHCL=2x (mol)

nCuO=y (mol) theo pt(2): nHCL=2y (mol)

Theo pt (1) và (2) ta có :

mCaO+mCuO =mhh hay 56x + 80y =1.36 (g) (*)

Mặt khác ta lại có : nHCL = 2x+2y=0.04 (mol) (**)

Gỉai (*) và (**) ta đc : x=0.01(mol) ; y=0.01(mol)

nCaO=x=0.01(mol) mCaO=0.01 .56=0.56(g)

nCuO=y=0.01(mol)mCuO=0.01 . 80 =0.8(g)

%CaO=41.18%

%CuO= 58.82%

b., Theo pthh (1) và (2) :nCaO=nCaCL2=x=0.01(mol)

nCuO=nCuCL2=y=0.01(mol)

mCaCL2=0.01.111=1.11(g)

mCuCL2=0.01 . 135=1.35 (g)

CuO +2HCl= CuCl2 +H2O
ZnO+2HCl= ZnCl2 +H2O
gọi x,y là mol của CuO, ZnO
80x + 81y = 12.1
2x+2y = 0.3
=> x=0.05 , y=0.1 => mCuO= 4 %CuO=4/12.1 m ZnO=8.1 =>%ZnO=8.1/12.1
nH2SO4=1/2nHCl=0.3/2 =0.15
mH2SO4=0.15x98=14.7g => mddH2SO4=14.7/20%=73.5g

n_H2 = \(\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Fe + 2HCl ---> FeCl₂ + H₂

0,15    0,3           0,15     0,15

m_Fe = 0,15.56 = 8,4 (g)

m_Fe2O3 = 24,4 - 8,4 = 16 (g)

n_Fe2O3 = \(\dfrac{16}{160}=0,1\left(mol\right)\)

%m_Fe = \(\dfrac{8,4}{24,4}=34,42\%\)

%m_Fe2O3 = \(100\%-34,42\%=65,58\%\)

Fe₂O₃ + 6HCl ---> 2FeCl₃ + 3H₂O

0,1            0,6          0,2           0,3

n_{HCl (ban đầu)} = 0,8.1,5 = 1,2 (mol)

n_{HCl (dư)} = 1,2 - 0,3 - 0,6 = 0,3 (mol)

=> \(\left\{{}\begin{matrix}C_{MFeCl_3}=\dfrac{0,2+0,15}{0,8}=0,4375M\\C_{MHCl\left(dư\right)}=\dfrac{0,3}{0,8}=0,375M\end{matrix}\right.\)

PTHH: 

FeCl₃ + 3NaOH ---> Fe(OH)₃ + 3NaCl

0,35        1,05

HCl + NaOH ---> NaCl + H₂O

0,3       0,3

=> \(V_{ddNaOH}=\dfrac{1,05+0,3}{1}=1,35\left(l\right)=1350\left(ml\right)\)

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

            \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

a) Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)=n_{Zn}\)

\(\Rightarrow\%m_{Zn}=\dfrac{0,4\cdot65}{36,2}\cdot100\%\approx71,23\%\) \(\Rightarrow\%m_{Al_2O_3}=28,77\%\)

c) Ta có: \(n_{Al_2O_3}=\dfrac{36,2-0,4\cdot65}{102}=0,1\left(mol\right)\)

Theo PTHH: \(n_{HCl}=2n_{Zn}+6n_{Al_2O_3}=1,4\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{1,4\cdot36,5}{10\%}=511\left(g\right)\) \(\Rightarrow V_{ddHCl}=\dfrac{511}{1,1}\approx464,5\left(ml\right)=0,4645\left(l\right)\)

c) Theo PTHH: \(\left\{{}\begin{matrix}n_{ZnCl_2}=0,4\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{ZnCl_2}}=\dfrac{0,4}{0,4645}\approx0,86\left(M\right)\\C_{M_{AlCl_3}}=\dfrac{0,2}{0,4645}\approx0,43\left(M\right)\end{matrix}\right.\)