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

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

Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{FeCl_2}=0,1\left(mol\right)=n_{H_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\\m_{FeCl_2}=0,1\cdot127=12,7\left(g\right)\\C_{M_{FeCl_2}}=\dfrac{0,1}{0,1}=1\left(M\right)\\C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\end{matrix}\right.\)

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

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

\(0.2......0.4........0.2.............0.2\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(m_{FeCl_2}=0.2\cdot127=25.4\left(g\right)\)

\(m_{HCl}=0.4\cdot36.5=14.6\left(g\right)\)

Câu c và câu d không liên quan tới dữ liệu đề bài cho !

Tiếp Quang Nhân:

c. Ta có: \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

d. Ko tính đc (Nồng độ mol thì đc)

Ta có: \(C_{M_{HCl}}=\dfrac{0,4}{200:1000}=2M\)

a, \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

 \(m_{HCl}=109,5.10\%=10,95\left(g\right)\Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

Mol:      0,1      0,3         0,1       0,15

Ta có: \(\dfrac{0,1}{2}=\dfrac{0,3}{6}\) ⇒ Al hết, HCl hết

\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)

b, \(m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)

c, m_{dd sau pứ} = 2,7 + 109,5 - 0,15.2 = 111,9 (g)

\(C\%_{ddAlCl_3}=\dfrac{13,35.100\%}{111,9}=11,93\%\)

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\n_{HCl}=\dfrac{109,5\cdot10\%}{36,5}=0,3\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,1}{2}=\dfrac{0,3}{6}\) \(\Rightarrow\) Al và HCl đều p/ứ hết

\(\Rightarrow\left\{{}\begin{matrix}n_{AlCl_3}=0,1\left(mol\right)\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=0,1\cdot133,5=13,35\left(g\right)\\V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Al}+m_{ddHCl}-m_{H_2}=111,9\left(g\right)\)

\(\Rightarrow C\%_{AlCl_3}=\dfrac{13,35}{111,9}\cdot100\%\approx11,93\%\)

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

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

\(m_{FeCl_2}=0,1.127=12,7g\\ b)V_{H_2}=0,1.22,4=2,24l\\ c)C_{M_{HCl}}=\dfrac{0,2}{0,2}=1M\)

Bài 14 : 

\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)

        1          2             1          1 

      0,15     0,3           0,15     0,15

a) \(n_{H2}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,15.22,4=3,36\left(l\right)\)

b) \(n_{HCl}=\dfrac{0,15.2}{1}=0,3\left(mol\right)\)

\(V_{ddHCl}=\dfrac{0,3}{2}=0,15\left(l\right)\)

c) \(n_{FeCl2}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)

\(C_{M_{FeCl2}}=\dfrac{0,15}{0,15}=1\left(M\right)\)

 Chúc bạn học tốt

`n_[Zn]=13/65=0,2(mol)`

`Zn + 2HCl -> ZnCl_2 + H_2 \uparrow`

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

`a)V_[H_2]=0,2.22,4=4,48(l)`

`b)C%_[HCl]=[0,4.36,5]/150 .100~~9,73%`