a) Al+6HCl—>AlCl3+H2

b) -Số mol của hiđro là:

nH2=VH2/22,4=13,44/22,4=0,6(mol) 

-PTHH:Al+6HCl—>AlCl3+ H2

-Theo PTHH:1-6-1-1

-Theo bài ra:0,6-3,6-0,6-0,6

Từ PTHH =>nAl=0,6mol

                     n HCl=0,6mol

-Khối lượng nhôm là:

mAl=nAl×MAl=0,6×27=16,2(g) 

-Khối lượng của axit là:

mHCl=nHCl×MHCl=0,6× 36,5=21,9(g) 

\(a.2Al+6HCl--->2AlCl_3+3H_2\uparrow\)

b. Ta có: \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

Theo PT: \(n_{Al}=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.0,6=0,4\left(mol\right)\)

\(\Rightarrow m_{Al}=0,4.27=10,8\left(g\right)\)

Theo PT: \(n_{HCl}=2.n_{H_2}=2.0,6=1,2\left(mol\right)\)

\(\Rightarrow m_{HCl}=1,2.36,5=43,8\left(g\right)\)

PT: \(FeO+H_2\underrightarrow{t^o}Fe+H_2O\)

Ta có: \(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)

a, Theo PT: \(n_{Fe}=n_{FeO}=0,1\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)

b, Theo PT: \(n_{H_2}=n_{FeO}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

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

THeo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{1}{15}\left(mol\right)\)

\(\Rightarrow m_{Al}=\dfrac{1}{15}.27=1,8\left(g\right)\)

a)

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

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

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

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

b) \(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

c)

PTHH: Zn + H₂SO₄ --> ZnSO₄ + H₂

                      0,3<----------------0,3

=> \(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)

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

PTHH:

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

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

\(V_{H_2}=0,3.22,4=6,72\left(l\right)\\ b,m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

c, PTHH:

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

           0,2<------------------0,2

\(m_{H_2SO_4}=0,2.98=19,6\left(g\right)\)

a)

\(n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)\\ Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O\\ n_{Fe_3O_4} = \dfrac{1}{3}n_{Fe} = \dfrac{1}{15}(mol)\\ \Rightarrow m_{Fe_3O_4} = \dfrac{1}{15}.232 = 15,467(gam)\)

b)

\(n_{H_2} = \dfrac{4}{3}n_{Fe} = \dfrac{4}{15}(mol)\\ \Rightarrow V_{H_2} =\dfrac{4}{15}.22,4 = 5,973(lít)\)

\(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PTHH: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)

...............1...............4..............3.....................

...............1/15.........4/15.........0,2..................

a. \(m_{Fe_3O_4}=n_{Fe_3O_4}\cdot M_{Fe_3O_4}=\dfrac{1}{15}\cdot232=\dfrac{232}{15}\left(g\right)\)

b. \(V_{H_2\left(ĐKTC\right)}=n_{H_2}\cdot22,4=\dfrac{4}{15}\cdot22,4=\dfrac{448}{75}\left(l\right)\)

Bạn tách ra từng câu nhé!

Bài 3.

\(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{36}{56}=0,6428mol\)

\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)

0,6428 ----- 0,4285           ( mol )

\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)

0,857                                                      0,4285    ( mol )

\(m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=0,857.158=135,406g\)

Bài 4.

a.\(n_{Al_2O_3}=\dfrac{m_{Al_2O_3}}{M_{Al_2O_3}}=\dfrac{51}{102}=0,5mol\)

\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)

  1       0,75            0,5     ( mol )

\(m_{Al}=n_{Al}.M_{Al}=1.27=27g\)

\(V_{O_2}=n_{O_2}.22,4=0,75.22,4=16,8l\)

b.\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)

       1,5                                                      0,75   ( mol )

\(m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=1,5.158=237g\)

\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)

  0,5                                0,75   ( mol )

\(m_{KClO_3}=n_{KClO_3}.M_{KClO_3}=0,5.122,5=61,25g\)

Bài 5.

\(n_{Mg}=\dfrac{m_{Mg}}{M_{Mg}}=\dfrac{14,4}{24}=0,6mol\)

\(n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{4,48}{22,4}=0,2mol\)

\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\)

0,6 <   0,2                         ( mol )

0,4      0,2               0,2      ( mol )

\(m_{MgO}=n_{MgO}.M_{MgO}=0,2.40=8g\)

Bài 6.

\(n_{Cu}=\dfrac{m_{Cu}}{M_{Cu}}=\dfrac{12,8}{64}=0,2mol\)

\(2Cu+O_2\rightarrow\left(t^o\right)2CuO\)

 0,2     0,1            0,2  ( mol )

\(V_{kk}=V_{O_2}.5=\left(0,1.22,4\right).5=2,24.5=11,2l\)

\(m_{CuO}=n_{CuO}.M_{CuO}=0,2.80=16g\)

PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

Ta có: \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

a, \(n_{Fe}=2n_{Fe_2O_3}=0,2\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,2.56=11,2\left(g\right)\)

b, \(n_{H_2}=3n_{Fe_2O_3}=0,3\left(mol\right)\)

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

c, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Theo PT: \(n_{Mg}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,3.24=7,2\left(g\right)\)

`4H_2 + Fe_3 O_4` $\xrightarrow{t^o}$ `3Fe + 4H_2 O`

`n_{Fe} = (33,6)/56 = 0,6 (mol)`

`a.`

Theo phương trình: `n_{Fe_3 O_4} = 1/3n_{Fe} = 0,2 (mol)`

`-> m_{Fe_3 O_4} = 0,2 . 232 = 46,4 (g)`

`b.`

Theo phương trình: `n_{H_2} = 4/3n_{Fe} = 0,8 (mol)`

`-> V_{H_2} = 0,8 . 22,4 = 17,92 (l)`

`c.`

`2H_2 O` $\xrightarrow{\text{điện phân}}$ `2H_2 + O_2`

Theo phương trình: `n_{H_2 O} = H_2 = 0,8 (mol)`

`-> m_{H_2 O} = 0,8 . 18 = 14,4 (g)`

a) Khối lượng Fe3­O4 cần dùng để điều chế 33,6 g Fe:

232 x 0,2 = 46,4 (g)

b) Thể tích khí cần dùng: 0,8 x 22,4 =17,92 (lít).

nAl=16,2/27= 0,6(mol)

a) PTHH: 4 Al +3 O2 -to-> 2 Al2O3

nO2= 3/4 . nAl=3/4 . 0,6= 0,45(mol)

=> V(O2,đktc)=0,45 x 22,4=10,08(l)

b) nAl2O3= nAl/2=0,6/2=0,3(mol)

=>mAl2O3=102. 0,3= 30,6(g)

c) 2KMnO4 -to-> K2MnO4 + MnO2 + O2

nKMnO4= 2.nO2=2. 0,45=0,9(mol)

=>mKMnO4= 158 x 0,9= 142,2(g)