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

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

0,2                      0,2          0,3    ( mol )

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

\(m_{AlCl_3}=n_{AlCl_3}.M_{AlCl_3}=0,2.133,5=26,7g\)

a)\(n_{H_2}=\dfrac{19,832}{24,79}=0,8\left(mol\right)\)

\(PTHH:Zn+H_2SO_4\xrightarrow[]{}ZnSO_4+H_2\)

tỉ lệ       :1        1               1              1

số mol  :0,8      0,8            0,8          0,8

\(m_{Zn}=0,8.65=52\left(g\right)\)

b)\(m_{H_2SO_4}=0.8.98=78,4\left(g\right)\)

c)\(m_{ZnSO_4}=0,8.161=128\left(g\right)\)

1/ Theo ĐLBT KL, có: m_KL + m_Cl2 = m_muối

⇒ m_Cl2 = 31,425 - 6,93 = 24,495 (g)

\(\Rightarrow n_{Cl_2}=\dfrac{24,495}{71}=0,345\left(mol\right)\)

\(\Rightarrow V_{Cl_2}=0,345.22,4=7,728\left(l\right)\)

2/ Giả sử: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\)

⇒ 24x + 27y = 6,93 (1)

Các quá trình:

\(Mg^0\rightarrow Mg^{+2}+2e\)

x_____________2x (mol)

\(Al^0\rightarrow Al^{+3}+3e\)

y___________3y (mol)

\(Cl_2^0+2e\rightarrow2Cl^-\)

0,345_0,69 (mol)

Theo ĐLBT mol e, có: 2x + 3y = 0,69 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,12\left(mol\right)\\y=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,12.24=2,88\left(g\right)\\m_{Al}=0,15.27=4,05\left(g\right)\end{matrix}\right.\)

Bạn tham khảo nhé!

\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

PTHH: 2Al + 6HCl ---> 2AlCl₃ + 3H₂

           0,3 ---------------> 0,3 -----> 0,45

\(\rightarrow\left\{{}\begin{matrix}V_{H_2}=0,45.22,4=10,08\left(l\right)\\m_{AlCl_3}=0,3.133,5=40,05\left(g\right)\end{matrix}\right.\)

\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\\ pthh:2Al+6HCl->2AlCl_3+3H_2\)
          0,3                        0,3            0,45 
=> \(V_{H_2}=0,45.22,4=10,08\left(L\right)\\ m_{AlCl_3}=0,3.133,5=40,05\left(G\right)\)

2Al +6HCl --- > 2AlCl3 +3H2

nAl = 8,1 / 27 = 0,3  (mol)

nH2 = 3/2 . 0,3 = 0,45 (mol)

VH2 = 0,45 . 22,4 = 10,08 (l)

nAlCl3 = n Al = 0,3 (mol)

mAlCl3 = 0,3 . 133,5 = 40,05 (g)

Giải:

a) Số mol Cl2 là:

nCl2 = V/22,4 = 4,8/22,4 ≃ 0,2 (mol)

PTHH: Cu + Cl2 -> CuCl2

---------x------x--------x----

PTHH: 2Al + 3Cl2 -> 2AlCl3

----------y------1,5y------y--

b) Gọi nCu = x (mol) và nAl = y (mol)

Ta có hệ phương trình:

\(\left\{{}\begin{matrix}mCu+mAl=5,9\left(g\right)\\nCl2=0,2\left(mol\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}64x+27y=5,9\left(g\right)\\x+1,5y=0,2\left(mol\right)\end{matrix}\right.\)

Giải hpt, ta được:

\(\left\{{}\begin{matrix}x=0,05\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)

Phần trăm khối lượng mỗi chất trong hỗn hợp ban đầu là:

%mCu = (n.M/mhh).100 = (0,05.64/5,9).100 ≃ 54,2 %

=> %mAl = 100 - 54,2 = 45,8 %

Vậy ...

Th1: R thuộc hóa trị I

\(2R+2HCl\rightarrow2RCl+H_2\)

Ta có: \(n_R=n_{RCl}\)

⇔ \(\dfrac{m}{2M_R}=\dfrac{m}{2M_{RCl}}\)

⇔ \(\dfrac{5,4}{2M_R}=\dfrac{26,7}{2\left(M_R+35,5\right)}\)

⇔ \(M_R=9\) (g/mol) (loại)

Th2: R thuộc hóa trị II

\(R+2HCl\rightarrow RCl_2+H_2\)

Ta có: \(n_R=n_{RCl_2}\)

⇔ \(\dfrac{m}{M_R}=\dfrac{m}{M_{RCl_2}}\)

⇔ \(\dfrac{5,4}{M_R}=\dfrac{26,7}{M_R+35,5.2}\)

⇔ \(M_R=18\) (g/mol) (loại)

Th3: R thuộc hóa trị III

\(2R+6HCl\rightarrow2RCl_3+3H_2\)

Ta có: \(n_R=n_{RCl_3}\)

⇔ \(\dfrac{m}{2M_R}=\dfrac{m}{2M_{RCl_3}}\)

⇔ \(\dfrac{5,4}{2M_R}=\dfrac{26,7}{2\left(M_R+35,5.3\right)}\)

⇔ \(M_R=27\) (g/mol) (nhận)

⇒ R là Nhôm (Al)

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

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

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

\(m_{HCl}=n.M=0,6.36,5=21,9\left(g\right)\)

\(m_{ddHCl}=\dfrac{m_{HCl}.100}{C\%}=\dfrac{21,9.100}{73}=30\left(g\right)\)

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

a. \(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

PTHH : Al₂O₃ + 6HCl -> 2AlCl₃ + 3H₂O

              0,1         0,6          0,2                      ( mol )

\(m_{HCl}=0,6.36,5=21,9\left(g\right)\)

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

b. 

PTHH : 3O₂ + 4Al -> 2Al₂O₃

              0,15                 0,1         ( mol) 

\(V_{O_2}=0,15.22,4=3,36\left(l\right)\)

a. \(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

PTHH : Al₂O₃ + 3HCl -> 2AlCl₃ + 3H₂O

              0,1         0,3          0,2                      ( mol )

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

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

b. 

PTHH : 3O₂ + 4Al -> 2Al₂O₃

              0,15                 0,1         ( mol) 

\(V_{O_2}=0,15.22,4=3,36\left(l\right)\)

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

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