\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)

\(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Ag}=y\left(mol\right)\end{matrix}\right.\Rightarrow27x+108y=4,2\left(1\right)\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,1                           0,05           0,15

\(\Rightarrow m_{Al}=0,1\cdot27=2,7g\)

\(\Rightarrow m_{Ag}=4,2-2,7=1,5g\)

a)\(\%m_{Al}=\dfrac{2,7}{4,2}\cdot100\%=64,28\%\)

\(\%m_{Ag}=100\%-64,28\%=35,72\%\)

b)\(m_{muối}=0,05\cdot342=17,1g\)

\(n_k=n_{H_2}=0,125\left(mol\right)\)

a,b, \(PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

.............0,125...0,125....................0,125...

\(\Rightarrow m_{Fe}=7\left(g\right)\)

Do Cu không phản ứng với H₂SO₄ .

\(\Rightarrow m_{Cu}=m_{hh}-m_{Fe}=10-7=3\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%Fe=70\%\\\%Cu=30\%\end{matrix}\right.\)

c, Có : \(m_{dd}=m_{Fe}+m_{ddH_2SO_4}-m_{H_2}=206,75\left(g\right)\)

\(\Rightarrow C\%=\dfrac{m_{H_2SO_4}}{m_{dd}}.100\%\approx5,925\%\)

\(n_{H_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(0.15.....0.3.......................0.15\)

\(m_{Mg}=0.15\cdot24=3.6\left(g\right)\)

\(m_{Cu}=10-3.6=6.4\left(g\right)\)

\(\%Mg=\dfrac{3.6}{10}\cdot100\%36\%\)

\(\%Cu=64\%\)

\(V_{dd_{HCl}}=\dfrac{0.3}{2}=0.15\left(l\right)\)

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

Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,3\left(mol\right)=n_{ZnCl_2}\\n_{HCl}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0.3\cdot65}{35,5}\cdot100\%\approx54,93\%\\\%m_{Cu}=45,07\%\\C\%_{HCl}=\dfrac{0,6\cdot36,5}{500}\cdot100\%=4,38\%\\m_{ZnCl_2}=0,3\cdot136=40,8\left(g\right)\end{matrix}\right.\)

Mặt khác: \(\left\{{}\begin{matrix}m_{Cu}=35,5-0,3\cdot65=16\left(g\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{dd}=m_{KL}+m_{ddHCl}-m_{Cu}-m_{H_2}=518,9\left(g\right)\)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{40,8}{518,9}\cdot100\%\approx7,86\%\)

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

                a_____3a______________\(\dfrac{3}{2}\)a                   (mol)

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

                b_____2b_____________b                        (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}27a+24b=7,8\\\dfrac{3}{2}a+b=\dfrac{8,96}{22,4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2\cdot27}{7,8}\cdot100\%\approx69,23\%\\\%m_{Mg}=30,77\%\\C_{M_{HCl}}=\dfrac{0,2\cdot3+0,1\cdot2}{0,2}=4\left(M\right)\end{matrix}\right.\) 

\(n_{H2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)

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

         1         2              1           1

         a       0,4           0,2           1a

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

           1          2            1          1

          b         0,3         0,15        1b

a) Gọi a là số mol của Mg

           b là số mol của Fe

\(m_{Mg}+m_{Fe}=13,2\left(g\right)\)

⇒ \(n_{Mg}.M_{Mg}+n_{Fe}.M_{Fe}=13,2g\)

 ⇒ 24a + 56b = 13,2g (1)

Theo phương trình : 1a + 1b = 0,35(2)

Từ(1),(2), ta có hệ phương trình : 

            24a + 56b = 13,2g

              1a + 1b = 0,35

               ⇒ \(\left\{{}\begin{matrix}a=0,2\\b=0,15\end{matrix}\right.\)

\(m_{Mg}=0,2.24=4,8\left(g\right)\)

\(m_{Fe}=0,15.56=8,4\left(g\right)\)

⁰/₀Mg = \(\dfrac{4,8.100}{13,2}=36,36\)⁰/₀

⁰/₀Fe = \(\dfrac{8,4.100}{13,2}=63,64\)⁰/₀

b) \(n_{HCl\left(tổng\right)}=0,4+0,3=0,7\left(mol\right)\)

200ml = 0,2l

\(C_{M_{ddHCl}}=\dfrac{0,7}{0,2}=3,5\left(M\right)\)

c) \(m_{muối.clorua}=\left(0,2.95\right)+\left(0,15.127\right)=38,05\left(g\right)\)

 Chúc bạn học tốt

C là tính tổng khối lượng nha

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

         0,1<-------------------------0,15

=> \(\%m_{Al}=\dfrac{0,1.27}{9,1}.100\%=29,67\%=>\%m_{Cu}=100\%-29,67\%=70,33\%\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Mà Cu k phản ứng hoàn toàn vs dd HCl

PTHH : 2Al + 6HCl -> 2AlCl₃ + 3H₂

0,1 mol ----------------------------- 0,15 mol

\(m_{Al}=0,1.27=2,7\left(g\right)\)

\(\%Al=\dfrac{2,7}{9,1}.100\%=29,6\%\)

\(\%Cu=100\%-29,6\%=70,4\%\)