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

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

PTHH: 2Al + 6HCl ---> 2AlCl₃ + 3H₂

           0,1                                    0,15

=> V_H2 = 0,15.22,4 = 3,36 (l)

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

PTHH: CuO + H₂ --t^o--> Cu + H₂O

LTL: \(0,4>0,15\rightarrow\) CuO dư

Theo pthh: \(n_{CuO\left(pư\right)}=n_{Cu}=n_{H_2}=0,15\left(mol\right)\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{0,15.64}{0,15.64+\left(0,4-0,15\right).80}=32,43\%\\\%m_{CuO}=100\%-32,43\%=67,57\%\end{matrix}\right.\)

a. \(2Al+6HCl\rightarrow2AlCl_3+3H_2\left(1\right)\)

b. \(n_{Al}=\frac{m}{M}=\frac{2,7}{27}=0,1mol\)

Theo phương trình `(1)` \(n_{H_2}=\frac{3}{2}.n_{Al}=\frac{3}{2}.0,1=0,15mol\)

\(\rightarrow V_{H_2\left(ĐKTC\right)}=n.22,4=0,15.22,4=3,36l\)

c. \(CuO+H_2\rightarrow^{t^o}Cu+H_2O\left(2\right)\)

\(n_{CuO}=\frac{m}{M}=\frac{32}{80}=0,4mol\)

Tỷ lệ \(\frac{0,4}{1}>\frac{0,15}{1}\)

`->CuO` dư

Theo phương trình `(2)` \(n_{Cu}=n_{H_2}=0,15mol\)

\(n_{CuO\left(pứ\right)}=n_{H_2}=0,15mol\)

\(\rightarrow n_{CuO\left(dư\right)}=0,4-0,15=0,25mol\)

\(m\left(g\right)\text{ chất rắn }\hept{\begin{cases}CuO_{dư}=0,25mol\\Cu=0,15mol\end{cases}}\)

\(\rightarrow m=0,15.64+0,25.80=29,6g\)

\(\%m_{CuO\left(dư\right)}=\frac{0,25.80.100}{29,6}\approx67,6\%\)

\(\%m_{Cu}=100\%-67,6\%=32,4\%\)

a. \(n_{Al}=\dfrac{2.7}{27}=0,1\left(mol\right)\)

PTHH : 2Al + 6HCl -> 2AlCl₃ + 3H₂

             0,1                                0,15

b. \(V_{H_2}=0,15.22,4=3,36\left(l\right)\)

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

PTHH : CuO + H₂ ---t^o---> Cu + H₂O

             0,15     0,15            0,15

Ta thấy : 0,4 > 0,15 =>  CuO dư , H₂ dư

\(\%CuO=\dfrac{0,15.80}{0,15.80+0,15.64}.100\%=55,55\%\)

\(\%Cu=100\%-55,55\%=44,45\%\)

$a\big)2Al+6HCl\to 2AlCl_3+3H_2$

$b\big)$

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

Theo PT: $n_{H_2}=\dfrac{3}{2}n_{Al}=0,3(mol)$

$\to V_{H_2(đktc)}=0,3.22,4=6,72(l)$

$c\big)$

Theo PT: $n_{AlCl_3}=n_{Al}=0,2(mol)$

$\to m_{AlCl_3}=0,2.133,5=26,7(g)$

`a)PTHH:`

`2Al + 6HCl -> 2AlCl_3 + 3H_2`

`0,2`     `0,6`                               `0,3`       `(mol)`

`n_[Al]=[5,4]/27=0,2(mol)`

`b)V_[H_2]=0,3.22,4=6,72(l)`

`c)m_[dd HCl]=[0,6.36,5]/10 . 100 =219(g)`

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

 2Al + 6HCl -> 2AlCl₃ + 3H₂

     0,2     0,6       0,2      0,3

=>VH2=0,3.22,4=6,72l

=>mHcl=0,6.36,5=21,9g

=>mdd=219g

a) n_Al = 43,2/27 = 1,6 mol

2Al + 6HCl → 2AlCl3 + 3H2

1,6                                 \(\dfrac{1,6\times3}{2}=2,4\)

→ n_H2 = 2,4 mol → V_H2 = 2,4 x 22, 4 = 53,76 lít

b) nCuO = 64/80 = 0,8 mol

nH2 = 2,4 mol

→ H2 dư, phương trình tính theo số mol của CuO

CuO + H2 → Cu + H2O

0,8        0,8       0,8    0,8

Chất rắn sau phản ứng có Cu

mCu = 0,8 x 64 = 51,2 gam

Zn+2HCl->ZnCl2+H2

0,05--------------------0,05

CuO+H2-to>Cu+H2O

          0,05----0,05

n Zn=\(\dfrac{3,25}{65}=0,05mol\)

=>n CuO=\(\dfrac{6}{80}=0,075mol\)

=>CuO dư

=>m Cu=0,05.64=3,2g

=>m CuO dư=0,025.80=2g

\(a,PTHH:\\ Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\left(1\right)\\ CuO+H_2\underrightarrow{t^o}Cu+H_2O\left(2\right)\\ b,n_{Zn}=\dfrac{3,25}{65}=0,05\left(mol\right)\\ Theo.pt\left(1\right):n_{H_2}=n_{Zn}=0,05\left(mol\right)\\ n_{CuO}=\dfrac{6}{80}=0,075\left(mol\right)\\ LTL.pt\left(2\right):0,075>0,05\Rightarrow CuO,dư\\ Theo.pt\left(2\right):n_{Cu}=n_{CuO\left(pư\right)}=n_{H_2}=0,05\left(mol\right)\\ m_{Cu}=0,05.64=3,2\left(g\right)\\ c,m_{CuO\left(dư\right)}=\left(0,075-0,05\right).80=2\left(g\right)\)

a.b.

\(n_{Fe}=\dfrac{8,4}{56}=0,15mol\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,15                              0,15  ( mol )

\(V_{H_2}=0,15.22,4=3,36l\)

c.\(n_{CuO}=\dfrac{9,6}{80}=0,12mol\)

\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)

0,12 < 0,15                              ( mol )

0,12                       0,12              ( mol )

\(m_{Cu}=0,12.64=7,68g\)

Help tui trong 5 phút sẽ đc điểm

Fe+2HCl->FeCl2+H2

0,15-------------------0,15

CuO+H2-to>Cu+H2O

0,12-------------0,12

n Fe=\(\dfrac{8,4}{56}\)=0,15 mol

=>VH2=0,15.22,4=3,36l

n CuO=\(\dfrac{9,6}{80}\)=0,12 mol

=>H2 dư

=>m Cu=0,12.64=7,68g

a) \(PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\) 

b) \(n_{Mg}=\dfrac{4,8}{24}=0,2mol\) 

\(n_{HCl}=2.n_{Mg}=0,2.2=0,4mol\)

\(\Rightarrow m_{HCl}=n.M=0,4.36,5=14,6g\)

c) \(n_{H_2}=n_{Mg}=0,2mol\) 

Thể tích khí hidro sinh ra (ở đktc): 

\(V_{H_2}=0,2.24,79=4,958l.\)