\(a,n_{Zn}=\dfrac{8,125}{65}=0,125\left(mol\right)\)

PTHH: Zn + 2HCl ---> ZnCl₂ + H₂

          0,125->0,25----->0,125->0,125

\(\Rightarrow\left\{{}\begin{matrix}a,V_{ddHCl}=\dfrac{0,25}{0,5}=0,5\left(l\right)\\b,V_{H_2}=0,125.22,4=2,8\left(l\right)\\c,C_{M\left(ZnCl_2\right)}=\dfrac{0,125}{0,5}=0,25M\end{matrix}\right.\)

a) Số mol kẽm tham gia phản ứng : \(n_{Zn}=\dfrac{m_{Zn}}{M_{Zn}}=\dfrac{16,25}{65}=0,25\left(mol\right)\).

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

Mol :       1    :     2     :       1      :   1

Mol :     0,25 → 0,5   →   0,25   →  0,5

Suy ra, số mol dung dịch Axit Clohidric \(HCl\) tham gia phản ứng là \(n_{HCl}=0,5\left(mol\right)\).

Khối lượng dung dịch đã dùng : \(m_{HCl}=n_{HCl}.M_{HCl}=\left(0,5\right).\left(36,5\right)=18,25\left(g\right)\).

b) Từ câu a, suy ra số mol khí Hidro sinh ra là \(n_{H_2}=0,25\left(mol\right)\).

Thể tích khí Hydro sinh ra là : \(V_{H_2}=n_{H_2}.22,4=\left(0,25\right).\left(22,4\right)=5,6\left(l\right)\)

\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) 
           0,4       0,8        0,4           0,4
\(a,V_{H_2}=0,4.22,4=8,96\left(l\right)\\ b,C\%_{HCl}=\dfrac{0,8.36,5}{150}.100\%=19,5\%\\ c,m_{\text{dd}}=26+150-\left(0,4.2\right)=175,2\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,4.136}{175,2}.100\%=31\%\)

\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)

\(m_{HCl}=\dfrac{300.10,95}{100}=32,85\left(g\right)\)

\(n_{HCl}=\dfrac{32,85}{36,5}=0,9\left(mol\right)\)

PTHH :

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

trc p/u:         0,4     0,9 

p/u :             0,4     0,8          0,4        0,4

sau p/u:         0      0,1          0,4         0,4 

---> sau p/ư : HCl dư 

\(a,V_{H_2}=0,4.22,4=8,96\left(l\right)\)

\(b,m_{HCl}=0,8.36,5=29,2\left(g\right)\)

\(c,m_{ZnCl_2}=0,4.136=54,4\left(g\right)\)

\(m_{ddZnCl_2}=26+300-\left(0,4.2\right)=325,2\left(g\right)\)

\(C\%_{ZnCl_2}=\dfrac{54,4}{325,2}.100\%\approx16,73\%\)

\(m_{HCldư}=0,1.36,5=3,65\left(g\right)\)

\(C\%_{HCldư}=\dfrac{3,65}{300}.100\%\approx1,22\%\)

a, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

PTHH: Zn + 2HCl ---> ZnCl₂ + H₂

           0,1--->0,2------->0,1----->0,1

V_H2 = 0,1.22,4 = 2,24 (l)

b, \(C_{M\left(HCl\right)}=\dfrac{0,2}{0,2}=1M\)

c, \(C_{M\left(ZnCl_2\right)}=\dfrac{0,1}{0,2}=0,5M\)

\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

\(m_{HCl}=\dfrac{150.18,25}{100}=27,375\left(g\right)\)

\(n_{HCl}=\dfrac{27,375}{36,5}=0,75\left(mol\right)\)

PTHH :

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

trc p/u :   0,3       0,75        

p/u:          0,3     0,6           0,3        0,3 

sau p/u :  0         0,15         0,3       0,3 

---> Sau p/ư HCl dư 

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

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

\(c,m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)

\(m_{ddZnCl_2}=19,5+150-\left(0,3.2\right)=168,9\left(g\right)\)

\(C\%=\dfrac{40,8}{168,9}.100\%\approx24,16\%\)

* Tóm tắt:

Biết: m_Zn = 6,5 (g)

        C_M(dd.HCl) = 0,5M

Hỏi: a) V_H2 = ?

       b) V_{dd HCl} = ?

a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

           0,1-->0,2------------->0,1

V_H2 = 0,1.22,4 = 2,24 (l)

b) \(V_{\left(dd.HCl\right)}=\dfrac{0,2}{0,5}=0,4\left(l\right)\)

  Tóm tắt:

`m_[Zn]=6,5 g`

`C_[M_[HCl]]=0,5 M`

________________________

`a)V_[H_2]=? l`

`b)V_[dd HCl]= ? l`

       Giải:

`Zn + 2HCl -> ZnCl_2 + H_2`

`0,1`    `0,2`                          `0,1`      `(mol)`

`n_[Zn]=[6,5]/65=0,1(mol)`

`a)V_[H_2]=0,1.22,4=2,24(l)`

`b)V_[dd HCl]=[0,2]/[0,5]=0,4(l)`

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(m_{HCl}=100.14,6\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\), ta được HCl dư.

Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{HCl\left(pư\right)}=2n_{Zn}=0,2\left(mol\right)\\n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\)

Ta có: m _{dd sau pư} = 6,5 + 100 - 0,1.2 = 106,3 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,1.136}{106,3}.100\%\approx12,79\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,2.36,5}{106,3}.100\%\approx6,87\%\end{matrix}\right.\)

\(a) 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ b)n_{AlCl_3} = n_{Al} = \dfrac{8,1}{27} = 0,3(mol)\\ \Rightarrow m_{AlCl_3} = 0,3.133,5 = 40,05(gam)\\ c) n_{HCl} = 3n_{Al} = 0,9(mol)\\ \Rightarrow C_{M_{HCl}} = \dfrac{0,9}{0,5} = 1,8M\)

a)\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1mol\)

b)Theo pt: \(n_{H_2}=n_{Zn}=0,1mol\Rightarrow V_{H_2}=0,1\cdot22,4=2,24l\)

c)\(Và\) \(n_{HCl}=2n_{Zn}=2\cdot0,1=0,2mol\)

\(\Rightarrow m_{HCl}=0,2\cdot36,5=7,3g\)

\(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%\Rightarrow m_{ddHCl}=\dfrac{m_{ct}}{C\%}\cdot100\%=\dfrac{7,3}{7,3\%}\cdot100\%=100g\)