nFe = 5,6 : 56 = 0,1(mol)
pthh : Fe + 2 HCl -->FeCl2 +  H2 
          0,1---------------> 0,1-----> 0,1 (mol) 
=> m = mFeCl2 = 0,1 .127 = 12,7 (g) 
=> V = VH2 (dktc ) = 0,1 . 22,4 = 2,24 (l) 
nCuO = 4 : 80 0,05 (mol) 
pthh CuO + H2 -t--> Cu + H2O
  LTL : 0,05/ 1   <  0,1 /1  => H2 du 
nH2(pu) = nCuO = 0,05 (mol) 
=> nH2 (du) = nH2 (ban dau ) - nH2 (pu ) 
                    = 0,1 - 0,05 = 0,05 (mol) 
mH2(du) = 0,05 . 2 = 0,1 (g)

Fe + 2HCl ---> FeCl₂ + H₂

0,1 -> 0,2   ->   0,1 ->  0,1 (mol)

n_Fe = \(\dfrac{5,6}{56}\)= 0,1 (mol)

m_FeCl2 = 0,1 . (56 + 35,5 . 2) = 12,7 (g)

VH2 = 0,1 . 22,4 = 2,24 (l)

b)                H₂ + CuO --> Cu + H₂O

                0,05 <- 0,05 -> 0,05 -> 0,05 (mol)

n_CuO = \(\dfrac{4}{80}\)= 0,05(mol)

Tỉ lệ : \(\dfrac{0,1}{1}\)   > \(\dfrac{0,05}{1}\). Vậy H₂ dư, tính theo CuO.

n_H2(dư) = n_{H2( ban đầu)} - n_{H2(phản ứng)} = 0,1 - 0,05 = 0,05 (mol)

Vui lòng kiểm tra lại, nếu có sai sót gì thì sorry.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

0,2                                              0,3    ( mol )

\(V_{H_2}=n_{H_2}.22,4=0,3.22,4=6,72l\)

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

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

 0,4  <  0,3                          ( mol )

0,3        0,3      0,3                   ( mol )

\(m_A=m_{CuO\left(du\right)}+m_{Cu}=\left[\left(0,4-0,3\right).80\right]+\left(0,3.64\right)=8+19,2=27,2g\)

:)) PTHH dưới em thiếu đk nhiệt độ

\(\left(a\right)2Al+3H_2O\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ \left(b\right)n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\\ n_{Al}=\dfrac{0,6.2}{3}=0,4mol\\ m_{Al}=0,4.27=10,8g\\ \left(c\right)n_{O_2}=\dfrac{4,8}{32}=0,15mol\\ 4Al+3O_2\underrightarrow{t^0}2Al_2O_3\\ \Rightarrow\dfrac{0,4}{4}>\dfrac{0,15}{3}\Rightarrow Al.dư\\ n_{Al_2O_3}=\dfrac{0,15.2}{3}=0,1mol\\ m_{oxit}=m_{Al_2O_3}=0,1.102=10,2g\)

a: \(2Al+3H_2SO_4\rightarrow1Al_2\left(SO_4\right)_3+3H_2\uparrow\)

    0,4           0,6                0,2             0,6

b: \(n_{H_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)

=>\(n_{Al}=0.4\left(mol\right)\)

\(m_{Al}=0.4\cdot27=10.8\left(g\right)\)

c: \(4Al+3O_2\rightarrow2Al_2O_3\)

0,4                      0,2

\(m_{Al_2O_3}=0.2\left(27\cdot2+16\cdot3\right)=0.2\cdot102=20.4\left(g\right)\)

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

b) \(n_{ZnCl_2}=\dfrac{20,4}{136}=0,15\left(mol\right)\)

\(n_{H_2}=n_{ZnCl_2}=0,15\left(mol\right)\)

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

c) \(m_{H_2}=0,15.2=0,3\left(g\right)\)

\(n_{HCl}=2n_{H_2}=0,3\left(mol\right)\)

\(m_{HCl}=0,3.36,5=10,95\left(g\right)\)

\(m_{Zn}=20,4+0,3-10,95=9,75\left(g\right)\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) 
           0,2                                 0,2 
\(V=V_{H_2}=0,2.22,4=4,48\left(l\right)\) 
\(pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\) 
                      0,2    0,2  
=> \(m_{Cu}=0,2.64=12,8\left(g\right)\)

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

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

_____0,2__________________0,2 (mol)

b, \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

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

___________0,2__0,2 (mol)

\(\Rightarrow m_{Cu}=0,2.64=12,8\left(g\right)\)

Bạn tham khảo nhé!

2Al+3H2SO4->Al2(SO4)3+3H2

0,1----------------------0,075----0,15

n H2=0,15 mol

=>mAl=0,1.27=2,7g

=>m Al2(SO4)3=0,075.342=25,65g

a) PTHH: \(2Al+3H_2SO_2\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(n_{Al}=\dfrac{2}{3}.0,15=0,1\left(mol\right)\)

\(m_{Al}=0,1.27=2,7\left(g\right)\)

c) \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)

\(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)

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