a) PT: Fe+2HCl→FeCl₂+H₂   (1) 

- Số mol Fe là:

n_Fe=\(\dfrac{m}{M}\)=\(\dfrac{11,2}{56}\)=0,2(mol)

- Theo PT (1)⇒n_FeCl2=n_Fe=0,2(mol)

- Vậy khối lượng của FeCl₂ là:

m_FeCl2=n.M=0,2.127=25,4(g)

b) Theo PT (1)⇒n_H2=n_Fe=0,2(mol)

- Vậy thể tích của H₂ là:

V_H2=n.24,79=0,2.24,79=4,958(l)

`#3107.101107`

`a)`

\(\text{Fe + 2HCl}\rightarrow\text{FeCl}_2+\text{H}_2\)

n của Fe có trong phản ứng là:

\(\text{n}_{\text{Fe}}=\dfrac{\text{m}_{\text{Fe}}}{\text{M}_{\text{Fe}}}=\dfrac{11,2}{56}=0,2\left(\text{mol}\right)\)

Theo PT: \(\text{n}_{\text{Fe}}=\text{n}_{\text{ }\text{FeCl}_2}=0,2\left(\text{mol}\right)\)

m của FeCl₂ có trong phản ứng là:

\(\text{m}_{\text{FeCl}_2}=\text{n}_{\text{FeCl}_2}\cdot\text{M}_{\text{FeCl}_2}=0,2\cdot\left(56+35,5\cdot2\right)=25,4\left(\text{g}\right)\)

`b)`

Theo PT: \(\text{n}_{\text{Fe}}=\text{n}_{\text{H}_2}=0,2\left(\text{mol}\right)\)

V của khí H₂ ở đkc là:

\(\text{V}_{\text{H}_2}=\text{n}_{\text{H}_2}\cdot24,79=0,2\cdot24,79=4,958\left(\text{l}\right)\)`.`

a, Ta có: 27n_Al + 56n_Fe = 22 (1)

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

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

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{19,832}{24,79}=0,8\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,4\left(mol\right)\\n_{Fe}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,4.27}{22}.100\%\approx49,09\%\\\%m_{Fe}\approx50,91\%\end{matrix}\right.\)

b, \(n_{HCl}=2n_{H_2}=1,6\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{1,6}{0,5}=3,2\left(M\right)\)

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

                   a_____2a______a_____a      (mol)

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

                    b_____3b_______b_____\(\dfrac{3}{2}\)b         (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}56a+27b=36,1\\a+\dfrac{3}{2}b=\dfrac{21,28}{22,4}=0,95\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,5\\b=0,3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,5\cdot56=28\left(g\right)\\m_{Al}=8,1\left(g\right)\end{matrix}\right.\)

b+c) Theo các PTHH: \(\left\{{}\begin{matrix}n_{HCl}=2a+3b=1,9\left(mol\right)\\n_{FeCl_2}=0,5\left(mol\right)\\n_{AlCl_3}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{HCl}}=\dfrac{1,9}{0,2}=9,5\left(M\right)\\C_{M_{FeCl_2}}=\dfrac{0,5}{0,2}=2,5\left(M\right)\\C_{M_{AlCl_3}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\end{matrix}\right.\)

\(a.BTNT\left(H\right):n_{HCl}=2n_{H_2}=0,65\left(mol\right)\\ \Rightarrow CM_{HCl}=\dfrac{0,65}{0,5}=1,3M\\ b.2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Đặt:\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+y=0,325\\27x+56y=9,65\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Al}=4,05\left(g\right)\\m_{Fe}=5,6\left(g\right)\end{matrix}\right.\)

\(n_{Fe}=\dfrac{6,72}{56}=0,12\left(mol\right)\\ Fe+H_2SO_{4\left(loãng\right)}\rightarrow FeSO_4+H_2\uparrow\\ Mol:0,12\rightarrow0,12\rightarrow0,12\rightarrow0,12\\ V_{H_2}=0,12.22,4=2,688\left(l\right)\\ m_{FeSO_4}=0,12.152=18,24\left(g\right)\\ PTHH:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\\ Mol:0,04\leftarrow0,12\rightarrow0,08\\ m_{Fe}=0,08.56=4,48\left(g\right)\)

Câu 5 : 

\(n_{H2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)

            1           2             1           1

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

         \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)

             1           2              1             1

           0,15       0,3          0,15

a) \(n_{Mg}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(m_{Mg}=0,1.24=2,4\left(g\right)\)

\(m_{MgO}=8,4-2,4=6\left(g\right)\)

⁰/₀Mg = \(\dfrac{2,4.100}{8,4}=28,57\)⁰/₀

⁰/₀MgO = \(\dfrac{6.100}{8,4}=71,43\)⁰/₀

b) Có : \(m_{MgO}=6\left(g\right)\)

\(n_{MgO}=\dfrac{6}{40}=0,15\left(mol\right)\)

\(n_{HCl\left(tổng\right)}=0,2+0,3=0,5\left(mol\right)\)

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

\(m_{ddHCl}=\dfrac{18,25.100}{3,65}=500\left(g\right)\)

\(n_{MgCl2\left(tổng\right)}=0,1+0,15=0,25\left(mol\right)\)

⇒ \(m_{MgCl2}=0,15.95=14,25\left(g\right)\)

\(m_{ddspu}=8,4+500-\left(0,1.2\right)=508,2\left(g\right)\)

\(C_{MgCl2}=\dfrac{14,25.100}{508,2}=2,8\)⁰/₀

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

Pt : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2|\)

        1           1               1           1

      0,1         0,1            0,1

a) \(n_{H2SO4}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(V_{ddH2SO4}=\dfrac{0,1}{2}=0,05\left(l\right)\)

b) \(n_{FeSO4}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(C_{M_{FeSO4}}=\dfrac{0,1}{0,05}=2\left(M\right)\)

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