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

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 \(\left(0,4-0,3\right)\cdot80=8g\)

\(m_{rắn}=m_{Cu}=0,3\cdot64=19,2g\)

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 \(\left\{{}\begin{matrix}Cu:0,3\left(mol\right)\\CuO\left(dư\right):0,1\left(mol\right)\end{matrix}\right.\)

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

Đổi 400ml = 0,4l

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂ (1)

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

Theo PTHH (1) \(n_{HCl}=3n_{Al}=3.0,2=0,6mol\)

→ CM ddHCl = 0,6/0,4 = 1,5M

Theo PTHH (1) nH2 = 3/2nAl = 3/2.0,2 = 0,3(mol)

nCuO = 32/80 = 0,4(mol)

PTHH: CuO + H2 —t°-> Cu + H2O

Trước pư: 0,4 0,3(mol)

Khi pư: 0,3 0,3 0,3(mol)

Sau pư: 0,1 0 0,3(mol)

mCuO dư = 0,1. 80 = 8(g)

mCu = 0,3. 64 = 19,2(g)

Trong m có 8gCuO dư và 19,2g Cu

%CuO = (8/27,2).100% = 29,4%; %Cu = 70,6%

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

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Theo PTHH: \(n_{Al}:n_{HCl}=2:6\)

\(\Rightarrow n_{HCl}=n_{Al}.3=0,2.3=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\frac{0,6}{0,4}=1,5\left(M\right)\)

b) Theo PTHH: \(n_{Al}:n_{H_2}=2:3\)

\(\Rightarrow n_{H_2}=n_{Al}.\frac{3}{2}=0,2.\frac{3}{2}=0,3\left(mol\right)\)

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

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

\(\left\{{}\begin{matrix}\frac{n_{H_2}}{1}=\frac{0,3}{1}=0,3\\\frac{n_{CuO}}{1}=\frac{0,4}{1}=0,4\end{matrix}\right.\) \(\Rightarrow\) CuO dư, H₂ phản ứng hết như vậy tính toán theo \(n_{H_2}\)

\(\Rightarrow n_{CuO\left(dư\right)}=n_{CuO\left(bđ\right)}-n_{CuO\left(pứ\right)}=0,4-0,3 =0,1\left(mol\right)\)

\(\Rightarrow\%m_{Cu}=\frac{0,3.64}{0,3.64+0,1.80}.100\%=70,59\%\)

\(\Rightarrow\%m_{CuO\left(dư\right)}=24,41\%\)

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

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Theo PT: \(n_{HCl}=3n_{Al}=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,4}=1,5\left(M\right)\)

b, \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)

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

Theo PT: \(n_{Cu}=n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Cu}=0,3.64=19,2\left(g\right)\)

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

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

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

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

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

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

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

=> CuO dư 

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

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

\(\%CuO\left(dư\right)=55.56\%\)

\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)

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

\(0.2.........................0.2.......0.3\)

\(m_{AlCl_3}=0.2\cdot133.5=26.7\left(g\right)\)

\(n_{CuO}=\dfrac{32}{160}=0.2\left(mol\right)\)

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

\(1..........1\)

\(0.2........0.3\)

\(LTL:\dfrac{0.2}{1}< \dfrac{0.3}{1}\Rightarrow H_2dư\)

\(n_{Cu}=0.2\left(mol\right)\)

\(m_{Cu}=0.2\cdot64=12.8\left(g\right)\)

Em xem lại đề vì chất rắn chỉ có Cu không có CuO nhé !

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 : \(0,2\left(mol\right)Cu\)

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

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

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