PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

Ta có: \(\left\{{}\begin{matrix}n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\n_{HCl}=\dfrac{109,5\cdot10\%}{36,5}=0,3\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\) \(\Rightarrow\) HCl còn dư, Magie p/ứ hết

\(\Rightarrow n_{Mg}=n_{MgCl_2}=n_{H_2}=0,1\left(mol\right)=n_{HCl\left(dư\right)}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,1\cdot95=9,5\left(g\right)\\m_{HCl\left(dư\right)}=0,1\cdot36,5=3,65\left(g\right)\\m_{H_2}=0,1\cdot2=0,2\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(saup/ứ\right)}=m_{Mg}+m_{ddHCl}-m_{H_2}=111,7\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{9,5}{111,7}\cdot100\%\approx8,5\%\\C\%_{HCl\left(dư\right)}=\dfrac{3,65}{111,7}\cdot100\%\approx3,27\%\end{matrix}\right.\)

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

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

0,1        0,3          0,1        0,15                ( mol )

\(m_{ddHCl}=\dfrac{0,3.36,5.100}{14,6}=75g\)

\(m_{ddspứ}=2,7+75-0,15.2=77,4g\)

\(C\%_{AlCl_3}=\dfrac{0,1.133,5}{77,4}.100=17,24\%\)

\(C\%_{H_2}=\dfrac{0,15.2}{77,4}.100=0,38\%\)

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

Ta lại có: \(C_{\%_{H_2SO_4}}=\dfrac{m_{H_2SO_4}}{294}.100\%=20\%\)

=> \(m_{H_2SO_4}=58,8\left(g\right)\)

=> \(n_{H_2SO_4}=\dfrac{58,8}{98}=0,6\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

Ta thấy: \(\dfrac{0,2}{1}=\dfrac{0,6}{3}\)

Vậy không có chất dư.

Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)

=> \(m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\)

Ta có: \(m_{dd_{Al_2\left(SO_4\right)_3}}=294+5,4-\left(\dfrac{3}{2}.0,2.2\right)=298,8\left(g\right)\)

=> \(C_{\%_{Al_2\left(SO_4\right)_3}}=\dfrac{34,2}{298,8}.100\%=11,45\%\)

\(n_{Zn}=\dfrac{65}{65}=1mol\)

\(n_{HCl}=\dfrac{200.14,6\%}{36,5}=0,8mol\)

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

 1    <   0,8                              ( mol )

0,4        0,8           0,4        0,4    ( mol )

\(m_{ddspứ}=200+65-0,4.2=264,2g\)

\(\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,4.136}{264,2}.100=14,93\%\\C\%_{H_2}=\dfrac{0,4.2}{264,2}.100=0,3\%\\C\%_{Zn\left(dư\right)}=\dfrac{\left(1-0,4\right).65}{264,2}.100=14,76\%\end{matrix}\right.\)

\(n_{Fe}=\dfrac{6,5}{56}=\dfrac{13}{112}mol\)

\(m_{CH_3COOH}=\dfrac{90\cdot20\%}{100\%}=18g\Rightarrow n_{CH_3COOH}=0,3mol\)

\(Fe+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Fe+H_2\uparrow\)

\(\dfrac{13}{112}\)       0,3                       0                        0

\(\dfrac{13}{112}\)       \(\dfrac{13}{56}\)                       \(\dfrac{13}{112}\)                   \(\dfrac{13}{112}\)

   0       \(\dfrac{19}{280}\)                      \(\dfrac{13}{112}\)                  \(\dfrac{13}{112}\)

a)\(m_{\left(CH_3COO\right)_2Fe}=\dfrac{13}{112}\cdot174=20,2g\)

\(m_{H_2}=\dfrac{13}{112}\cdot2=\dfrac{13}{56}g\)

\(m_{dd\left(CH_3COO\right)_2Fe}=6,5+90-\dfrac{13}{56}=96,27g\)

\(C\%=\dfrac{20,2}{96,27}\cdot100\%=20,98\%\)

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(n_{HCl}=\dfrac{500.14,6\%}{36,5}=2\left(mol\right)\)

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

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{2}{6}\), ta được HCl dư.

Theo PT: \(\left\{{}\begin{matrix}n_{HCl\left(pư\right)}=3n_{Al}=0,6\left(mol\right)\\n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\end{matrix}\right.\)

⇒ n_{HCl (dư)} = 2 - 0,6 = 1,4 (mol)

Ta có: m _{dd sau pư} = 5,4 + 500 - 0,3.2 = 504,8 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{1,4.36,5}{504,8}.100\%\approx10,12\%\\C\%_{AlCl_3}=\dfrac{0,2.133,5}{504,8}.100\%\approx5,29\%\end{matrix}\right.\)

\(n_{H2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

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

        2         6               2             3

        a       0,6             0,2          1,5a

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

         1          2              1          1

        b         0,2            0,1              1b

a) Gọi a là số mol của Al

           b là số mol của Fe

\(m_{Al}+m_{Fe}=11\left(g\right)\)

⇒ \(n_{Al}.M_{Al}+n_{Fe}.M_{Fe}=11g\)

 ⇒ 27a + 56b = 11g (1)

Theo phương trình : 1,5a + 1b = 0,4(2)

Từ (1),(2), ta có hệ phương trình : 

            27a + 56b = 0,4

            1,5a + 1b = 0,4

            ⇒ \(\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)

 \(m_{Al}=0,2.27=5,4\left(g\right)\)

\(m_{Fe}=0,1.56=5,6\left(g\right)\)

⁰/₀Al = \(\dfrac{5,4.100}{11}=49,09\)⁰/₀

⁰/₀Fe = \(\dfrac{5,6.100}{11}=50,91\)⁰/₀

b) \(n_{HCl\left(tổng\right)}=0,6+0,2=0,8\left(mol\right)\)

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

\(m_{ddHCl}=\dfrac{29,2.100}{20}=146\left(g\right)\)

c) \(n_{AlCl3}=\dfrac{0,6.2}{6}=0,2\left(mol\right)\)

⇒ \(m_{AlCl3}=0,2.133,5=26,7\left(g\right)\)

\(n_{FeCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

⇒ \(m_{FeCl2}=0,1.127=12,7\left(g\right)\)

\(m_{ddspu}=11+146-\left(0,4.2\right)=156,2\left(g\right)\)

\(C_{AlCl3}=\dfrac{26,7.100}{156,2}=17,09\)⁰/₀

\(C_{FeCl2}=\dfrac{12,7.100}{156,2}=8,13\)⁰/₀

 Chúc bạn học tốt