\(n_P=\dfrac{12}{31}mol\)

\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

\(\dfrac{12}{21}\)\(\rightarrow\)\(\dfrac{5}{7}\)

Khối lượng \(KMnO_4\) cần dùng để thu lượng khí trên:

\(2KMnO_4\underrightarrow{t^o}K_2MnO_2+MnO_2+O_2\)

\(\dfrac{10}{7}\)          \(\leftarrow\)                                    \(\dfrac{5}{7}\)

\(\Rightarrow m_{KMnO_4}=\dfrac{10}{7}\cdot158=225,71g\)

Nếu không dùng \(KMnO_4\) mà dùng \(KClO_3\) để thu được khí đủ dùng trên:

\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

\(\dfrac{10}{21}\)        \(\leftarrow\)               \(\dfrac{5}{7}\)

\(m_{KClO_3}=\dfrac{10}{21}\cdot122,5=58,33g\)

nP = 12/31 (mol)

PTHH: 4P + 5O2 -> (to) 2P2O5

Mol: 12/31 ---> 15/31

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

nKMnO4 = 15/31 . 2 = 30/31 (mol)

mKMnO4 = 30/31 . 158 = 4740/31 (g)

PTHH: 2KClO3 -> (to, MnO2) 2KCl + 3O2

nKClO3 = 30/31 : 3 . 2 = 15/31 (mol)

mKClO3 = 15/31 . 122,5 = 3675/62 (g)

`n_[CuO]=[0,8]/80=0,01(mol)`

`H_2 + CuO` $\xrightarrow{t^o}$ `Cu + H_2 O`

`0,01`   `0,01`        `0,01`                    `(mol)`

`a)m_[Cu]=0,01.64=0,64(g)`

`b)V_[H_2]=0,01.22,4=0,224(l)`

`c)`

`Fe + 2HCl -> FeCl_2 + H_2`

`0,01`  `0,02`                       `0,01`       `(mol)`

`@m_[Fe]=0,01.56=0,56(g)`

`@m_[dd HCl]=[0,02.36,5]/20 . 100=3,65(g)`

Chọn B

Đặt x,y lần lượt là số mol của N a 2 C O 3 và  (x,y>0)

N a 2 C O 3 +  H 2 S O 4   → N a 2 S O 4  + C O 2 + H₂O (1)

x mol           x mol          x mol      x mol

K 2 C O 3 +  H 2 S O 4   →    K 2 S O 4         +   C O 2      +    H 2 O  (2)

y mol         y mol           y mol               y mol

Từ phương trình ta dễ thấy muối cacbonat tác dụng với H 2 S O 4 thì

n H 2 S O 4 = n C O 2

\(a,n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ PTHH:2Al+3H_2SO_{4\left(loãng\right)}\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\\ Theo.pt:n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,4=0,6\left(mol\right)\\ b,PTHH:RO+H_2\underrightarrow{t^o}R+H_2O\\ Mol:0,6\leftarrow0,6\rightarrow0,6\\ M_R=\dfrac{38,4}{0,6}=64\left(\dfrac{g}{mol}\right)\\ \Rightarrow R.là.Cu\)

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

0,4---------------------------------0,6

n Al=0,4 mol

=>VH2=0,6.22,4=13,44l

b)

H2+XO-to>X+H2O

0,6------------0,6

=>0,6=\(\dfrac{38,4}{X}\)

=>X=64 đvC

=>X là Cu(đồng)

=>X=48

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

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

n Al=0,2 mol

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

b)

XO+H2-to>X+H2O

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

=>0,3=\(\dfrac{19,5}{X}\)

=>X là Zn( kẽm)

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

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

0,2                                    0,3  ( mol )

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

b.\(n_X=\dfrac{19,5}{M_X}\)

\(XO+H_2\rightarrow\left(t^o\right)X+H_2O\)

          \(\dfrac{19,5}{M_X}\)        \(\dfrac{19,5}{M_X}\)         ( mol )

Ta có:

\(\dfrac{19,5}{M_X}=0,3\)

\(\Leftrightarrow M_X=65\)

=> X là kẽm (Zn)

\(a,n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow AlCl_3+3H_2\uparrow\\ Theo.pt:n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ V_{H_2}=0,3.22,4=6,72\left(l\right)\\ b,PTHH:RO+H_2\underrightarrow{t^o}R+H_2O\\ Mol:0,3\leftarrow0,3\rightarrow0,3\\ M_R=\dfrac{19,5}{0,3}=65\left(\dfrac{g}{mol}\right)\\ \Rightarrow R.là.Zn\)

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

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

0,2                                         0,3    ( mol )

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

b.\(XO+H_2\rightarrow\left(t^o\right)X+H_2O\)

\(n_X=\dfrac{19,5}{M_X}\) mol

\(n_{H_2}=n_X=0,3mol\)

\(\Rightarrow\dfrac{19,5}{M_X}=0,3\)

\(M_X=65\) ( g/mol )

=> X là kẽm ( Zn )

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

PTHH: 2Al + 6HCl ---> 2AlCl₃ + 3H₂

           0,2      0,6             0,2      0,3

V_H2 = 0,3.22,4 = 6,72 (l)

b, PTHH: RO + H₂ ---t^o---> R + H₂O

                          0,3             0,3

=> M_R = \(\dfrac{19,5}{0,3}=65\left(\dfrac{g}{mol}\right)\)

=> R là Zn

Ta có: \(n_{Fe_2O_3}=\dfrac{48}{160}=0,3\left(mol\right)\)

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

_____0,3_____0,9___0,6____0,9 (mol)

a, \(m_{Fe}=0,6.56=33,6\left(g\right)\)

b, \(V_{H_2}=0,9.22,4=20,16\left(l\right)\)

c, PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{H_2O}=0,9\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,9.22,4=20,16\left(l\right)\)

Câu 2: 

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\)

               0,6<------------------------------------0,3

\(\Rightarrow m_{KMnO_4}=0,6.158=94,8\left(g\right)\)

Câu 3: 

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

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

             0,2<-----------------------0,2

=> m_Zn = 0,2.65 = 13 (g)

Câu 4:

\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)

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

           0,4------------------------->0,4

             \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

              0,4<---0,4

\(\Rightarrow m_{CuO}=0,4.80=32\left(g\right)\)

Mg+2HCl->MgCl2+H2

0,2-----------------------0,2

RO+H2-to>R+H2O

0,2-------------0,2

n Mg=\(\dfrac{4,8}{24}\)=0,2 mol

=>VH2=0,2.22,4=4,48l

->0,2=\(\dfrac{12,8}{R}\)

=>R=64 g\mol

=>R là Cu(đồng)