\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)

a)\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

    0,2     0,6                         0,3

\(C_M=\dfrac{0,6}{0,4}=1,5M\)

b)\(n_{CuO}=\dfrac{32}{80}=0,4mol\)

\(CuO+H_2\rightarrow Cu+H_2O\)

0,4        0,3     0,3

Sau phản ứng CuO dư và dư \(\left(0,4-0,3\right)\cdot80=8g\)

\(m_{rắn}=m_{Cu}=0,3\cdot64=19,2g\)

nAl = 5.4/27 = 0.2 (mol) 

2Al + 6HCl => 2AlCl3 + 3H2 

0.2.......0.6......................0.3

CM HCl = 0.6 / 0.4 = 1.5 (M) 

nCuO = 32/80 = 0.4 (mol) 

CuO + H2 -to-> Cu + H2O 

0.2.......0.2..........0.2 

Chất rắn : 0.2 (mol) CuO dư , 0.2 (mol) Cu 

%CuO = 0.2*80 / ( 0.2*80 + 0.2*64) * 100% = 55.56%

%Cu = 44.44%

\(n_{Mg}=\dfrac{4.8}{24}=0.2\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(0.2.......0.4........................0.2\)

\(C_{M_{HCl}}=\dfrac{0.4}{0.2}=2\left(M\right)\)

\(n_{CuO}=\dfrac{32}{80}=0.4\left(mol\right)\)

\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)

Lập tỉ lệ : \(\dfrac{0.4}{1}>\dfrac{0.2}{1}\)

=> CuO dư 

\(m_{cr}=m_{CuO\left(dư\right)}+m_{Cu}=32-0.2\cdot80+0.2\cdot64=28.8\left(g\right)\)

\(\%Cu=\dfrac{0.2\cdot64}{28.8}\cdot100\%=44.44\%\)

\(\%CuO\left(dư\right)=55.56\%\)

a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

            0,2---------------------->0,3

=> V_H2 = 0,3.22,4 = 6,72 (l)

b)

\(n_{CuO}=\dfrac{32}{80}=0,4\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

Xét tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,3}{1}\) => CuO dư, H₂ hết

PTHH: CuO + H₂ --t^o--> Cu + H₂O

              0,3<--0,3------->0,3

=> Rắn sau pư gồm \(\left\{{}\begin{matrix}Cu:0,3\left(mol\right)\\CuO\left(dư\right):0,1\left(mol\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{0,3.64}{0,3.64+0,1.80}.100\%=70,59\%\\\%m_{CuO}=\dfrac{0,1.80}{0,3.64+0,1.80}.100\%=29,41\%\end{matrix}\right.\)

a)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            0,2--->0,4---->0,2--->0,2

\(V_2=0,2.22,4=4,48\left(l\right)\)

\(V_1=\dfrac{0,4}{0,5}=0,8\left(l\right)\)

b)

\(C_{M\left(ZnCl_2\right)}=\dfrac{0,2}{0,8}=0,25M\)

c)

\(n_{H_2}=0,1\left(mol\right)\); \(n_{CuO}=\dfrac{32}{80}=0,4\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

Xét tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,1}{1}\) => CuO dư, H₂ hết

PTHH: CuO + H₂ --t^o--> Cu + H₂O

              0,1<--0,1------>0,1

=> m = 32 - 0,1.80 + 0,1.64 = 30,4 (g)

a) 

b) nH2=32nAl=0,3(mol)nH2=32nAl=0,3(mol)

→VH2(đktc)=0,3.22,4=6,72(l)→VH2(đktc)=0,3.22,4=6,72(l)

c) Chất rắn : 

CuO dư : 

a)\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

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

b)\(V_{H_2}=0,3\cdot22,4=6,72l\)

c)\(n_{CuO}=\dfrac{32}{80}=0,4mol\)

\(CuO+H_2\rightarrow Cu+H_2O\)

0,4       0,3       0,3

\(m_{Cu}=0,3\cdot64=19,2g\)

nAl = 5.4/27 = 0.2 (mol) 

2Al + 6HCl => 2AlCl3 + 3H2 

0.2.......0.6......................0.3

CM HCl = 0.6 / 0.4 = 1.5 (M) 

nCuO = 32/80 = 0.4 (mol) 

CuO + H2 -to-> Cu + H2O 

0.2.......0.2..........0.2 

Chất rắn : 0.2 (mol) CuO dư , 0.2 (mol) Cu 

%CuO =\(\dfrac{0,2.80}{0,2.80+0,2.64}\) 100% = 55.56%

%Cu = 44.44%

a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

Theo phương trình : \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)

\(\rightarrow V_{H_2}\left(đktc\right)=0,3.22,4=6,72\left(l\right)\)

c) Chất rắn : \(0,2\left(mol\right)\)

CuO dư : \(0,2\left(mol\right)Cu\)

\(\%CuO=\dfrac{0,2.80}{\left(0,2.80+0,2.64\right)}.100=55,56\%\)

\(\%Cu=44,44\%\)

111111111111111111111111111111111111111111111111ξ11111111111111

a) $Fe + 2HCl \to FeCl_2 + H_2$

 Theo PTHH : n H2 = n Fe = 8,4/56 = 0,15(mol)

V H2 = 0,15.22,4 = 3,36(lít)

b) n HCl = 2n Fe = 0,3(mol)

=> CM HCl = 0,3/0,2 = 1,5M

c) $CuO + H_2 \xrightarrow{t^o} Cu + H_2O$

Ta thấy :

n CuO = 32/80 = 0,4 > n H2 = 0,15 mol nên CuO dư

Theo PTHH : n Cu = n H2 = 0,15 mol

=> m Cu = 0,15.64 = 9,6 gam