\(a,n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

0,1      0,2             0,1          0,1

\(V_{H_2}=0,1.22,4=2,24l\\ b)C_{\%HCl}=\dfrac{0,2.36,5}{200}\cdot100=3,65g\)     

a.b.\(n_{Zn}=\dfrac{1,95}{65}=0,03mol\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,03   0,06          0,03           0,03    ( mol )

\(V_{H_2}=0,03.22,4=0,672l\)

\(m_{ddHCl}=\dfrac{0,06.36,5}{7,3\%}=30g\)

c.Tên muối: Kẽm clorua

\(m_{ZnCl_2}=0,03.136=4,08g\)

\(m_{ddspứ}=30+1,95-0,03.2=31,89g\)

\(C\%_{ZnCl_2}=\dfrac{4,08}{31,89}.100\%=12,79\%\)

nAl= 0,5(mol)

a) PTHH: 2 Al + 6 HCl -> 2 AlCl3 + 3 H2

nHCl= 6/2 . 0,5= 1,5(mol)

=>mHCl= 1,5.36,5=54,75(mol)

=> mddHCl= (54,75.100)/18,25=300(g)

b) nH2= 3/2. 0,5=0,75(mol)

=>V(H2,đktc)=0,75.22,4=16,8(l)

c) nAlCl3= nAl= 0,5(mol) -> mAlCl3=0,5. 133,5=66,75(g)

mddAlCl3=mAl+ mddHCl - mH2= 13,5 + 300-0,75.2=312(g)

=> \(C\%ddAlCl3=\dfrac{66,75}{312}.100\approx21,394\%\)

a) \(n_{HCl}=\dfrac{m}{M}=\dfrac{13}{65}=0,2\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) 
   1       2           1        1    (mol)

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

b) Thể tích khí hidro:

V = n.22,4 = 0,2.22,4 = 4,48 (l)

c) Khối lượng muối tạo thành:

\(m_{ZnCl_2}=n.M=0,2.\left(65+35,5.2\right)=27,2\left(g\right)\)

d) \(m_{ctHCl}=n.M=0,4.\left(1+35,5\right)=14,6\left(g\right)\)

\(C\%_{HCl}=\dfrac{m_{ctHCl}}{m_{ddHCl}}.100\%=\dfrac{14,6}{200}.100\%=7,3\%\)

a) Zn + 2HCl → ZnCl₂ + H₂↑

b) 

Câu 1:

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)\)

\(n_{H_2}=n_{Zn}=0,25\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,25.24,79=6,1975\left(l\right)\)

c, \(n_{HCl}=2n_{Zn}=0,5\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{18,25}{10\%}=182,5\left(g\right)\)

d, \(n_{ZnCl_2}=n_{Zn}=0,25\left(mol\right)\)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{0,25.136}{16,25+182,5-0,25.2}.100\%\approx17,15\%\)

Câu 2:

a, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

c, \(n_{NaOH}=\dfrac{40}{40}=1\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,5\left(mol\right)\)

\(\Rightarrow V_{H_2SO_4}=\dfrac{0,5}{2}=0,25\left(l\right)\)

d, \(n_{Na_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,5\left(mol\right)\)

\(\Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,5}{0,25}=2\left(M\right)\)

\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

a) Pt : \(CuO+2HCl\rightarrow CuCl_2+H_2O|\)

             1            2             1            1

             0,1        0,2           0,1

b) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

⇒ \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

\(C_{ddHCl}=\dfrac{7,3.100}{200}=3,65\)⁰/₀

c) \(n_{CuCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

⇒ \(m_{CuCl2}=0,1.135=13,5\left(g\right)\)

 Chúc bạn học tốt

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

a.

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

\(V_{H_2}=24,79.0,2=4,958\left(l\right)\)

b.

\(n_{HCl}=2.n_{Fe}=0,4\left(mol\right)\\ CM_{HCl}=\dfrac{0,4}{0,2}=2M\)

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

\(n_{H_2}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\)

Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,4\left(mol\right)\Rightarrow m_{Al}=0,4.27=10,8\left(g\right)\)

b, \(n_{HCl}=2n_{H_2}=1,2\left(mol\right)\Rightarrow C\%_{HCl}=\dfrac{1,2.36,5}{250}.100\%=17,52\%\)

c, m _{dd sau pư} = 10,8 + 250 - 0,6.2 = 259,6 (g)

d, \(n_{AlCl_3}=\dfrac{2}{3}n_{H_2}=0,4\left(mol\right)\)

\(\Rightarrow C\%_{AlCl_3}=\dfrac{0,4.133,5}{259,6}.100\%\approx20,57\%\)

\(n_{Al}=\dfrac{21,6}{27}=0,8\left(mol\right)\)

Pt : \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O|\)

        2             3                  1                3

       0,8          1,2               0,4             1,2

a) \(n_{H2}=\dfrac{0,8.3}{2}=1,2\left(mol\right)\)

\(V_{H2\left(dktc\right)}=1,2.22,4=26,88\left(l\right)\)

b) \(n_{H2SO4}=\dfrac{0,8.3}{2}=1,2\left(mol\right)\)

⇒ \(m_{H2SO4}=1,2.98=117,6\left(g\right)\)

\(m_{ddH2SO4}=\dfrac{117,6.100}{29,4}=400\left(g\right)\)

c) \(n_{Al2\left(SO4\right)3}=\dfrac{1,2.1}{3}=0,4\left(mol\right)\)

⇒ \(m_{Al2\left(SO4\right)3}=0,4.342=136,8\left(g\right)\)

\(m_{ddspu}=21,6+400-\left(1,2.2\right)=419,2\left(g\right)\)

\(C_{Al2\left(SO4\right)3}=\dfrac{136,8.100}{419,2}=32,63\)⁰/₀

 Chúc bạn học tốt