a) Gọi số mol Mg, Fe là a, b (mol)

=> 24a + 56b = 11,84

\(n_{HCl}=\dfrac{146.14\%}{36,5}=0,56\left(mol\right)\)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

            a--->2a--------->a----->a

           Fe + 2HCl --> FeCl₂ + H₂

            b-->2b-------->b------>b

=> 2a + 2b = 0,56

=> a = 0,12; b = 0,16

=> \(\left\{{}\begin{matrix}\%Mg=\dfrac{0,12.24}{11,84}.100\%=24,324\%\\\%Fe=\dfrac{0,16.56}{11,84}.100\%=75,676\%\end{matrix}\right.\)

b) \(n_{H_2}=a+b=0,28\left(mol\right)\)

=> \(V_{H_2}=0,28.22,4=6,272\left(l\right)\)

c) m_{dd sau pư} = 11,84 + 146 - 0,28.2 = 157,28 (g)

=> \(\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,12.95}{157,28}.100\%=7,25\%\\C\%_{FeCl_2}=\dfrac{0,16.127}{157,28}.100\%=12,92\%\end{matrix}\right.\)

a) \(n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

          0,05<-----------0,05---->0,075

=> \(\%Al=\dfrac{0,05.27}{14,15}.100\%=9,54\%\)

=> \(\%Cu=\dfrac{14,15-0,05.27}{14,15}.100\%=90,46\%\)

b) \(V_{H_2}=0,075.22,4=1,68\left(l\right)\)

c) \(n_{Cu}=\dfrac{14,15-0,05.27}{64}=0,2\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

          0,05->0,0375

           2Cu + O₂ --t^o--> 2CuO 

            0,2-->0,1

=> \(V_{O_2}=\left(0,1+0,0375\right).22,4=3,08\left(l\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ m_{AlCl_3}=6,675\left(mol\right)\\ n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\\ \Rightarrow n_{Al}=n_{AlCl_3}=0,05\left(mol\right)\\ \Rightarrow m_A=0,05.27=1,35\left(g\right);m_{Cu}=14,15-1,35=12,8\left(g\right)\\ \%m_{Cu}=\dfrac{12,8}{14,15}.100\approx90,459\%\\ \Rightarrow\%m_{Al}\approx9,541\%\\ b,n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,05=0,075\left(mol\right)\\ \Rightarrow V=V_{H_2\left(đktc\right)}=0,075.22,4=1,68\left(l\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ 2Cu+O_2\rightarrow\left(t^o\right)2CuO\\ n_{O_2}=\dfrac{3}{4}.n_{Al}+\dfrac{1}{2}.n_{Cu}=\dfrac{3}{4}.0,05+\dfrac{1}{2}.0,2=0,0875\left(mol\right)\)

\(\Rightarrow V_{O_2\left(đktc\right)}=0,0875.22,4=1,96\left(l\right)\)

\(n_{O_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

\(n_{H_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)

\(n_{Fe}=a\left(mol\right),n_{Al}=b\left(mol\right)\)

\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)

\(a.......\dfrac{2a}{3}\)

\(4Al+3O_2\underrightarrow{^{^{t^0}}}2Al_2O_3\)

\(b.......\dfrac{3b}{4}\)

\(n_{O_2}=\dfrac{2a}{3}+\dfrac{3b}{4}=0.25\left(mol\right)\left(1\right)\)

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

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

\(n_{H_2}=a+1.5b=0.45\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.15,b=0.2\)

\(m_{Fe}=0.15\cdot56=8.4\left(g\right)\)

\(m_{Al}=0.2\cdot27=5.4\left(g\right)\)

\(\%m_{Fe}=\dfrac{8.4}{8.4+5.4}\cdot100\%=60.8\%\)

\(\%m_{Al}=100-60.8=39.2\%\)

\(\left\{{}\begin{matrix}Zn\\Fe\\Mg\end{matrix}\right.+H_2SO_4\rightarrow\left\{{}\begin{matrix}ZnSO_4\\FeSO_4\\MgSO_4\end{matrix}\right.+H_2\uparrow\)

Ta có: \(m_{SO_4}=8,25-2,49=5,76\left(g\right)\)

\(\Rightarrow n_{H_2SO_4}=n_{H_2}=n_{SO_4}=\dfrac{5,76}{96}=0,06\left(mol\right)\)

a, \(m_{H_2SO_4}=0,06.98=5,88\left(g\right)\)

b, \(V_{H_2}=0,06.22,4=1,344\)

\(n_{HCl}=0,3.1=0,3mol\)

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

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

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

\(\Rightarrow m_{Al}=9,1.2,7=6,4g\)

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

nHCl = 0,3 . 1 = 0,3 (mol)

PTHH: 2Al + 6HCl -> 2AlCl3 + 3H2

Mol: 0,1 <--- 0,3 ---> 0,1 ---> 0,15

mAl = 0,1 . 27 = 2,7 (g(

mCu = 9,1 - 2,7 = 6,4 (g)

VH2 = 0,15 . 22,4 = 3,36 (l)

\(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\\n_{Fe}=c\left(mol\right)\end{matrix}\right.\)⇒ 24a + 27b + 56c = 26,05(1)

\(Mg + 2HCl \to MgCl_2 + H_2\\ 2Al +6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = a + 1,5b + c = \dfrac{13,44}{22,4} = 0,6(2)\)

\(Mg + Cl_2 \xrightarrow{t^o} MgCl_2\\ 2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3\\ 2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3\\ n_{Cl_2} = a + 1,5b + 1,5c = \dfrac{17,36}{22,4} = 0,775(3)\)

Từ (1)(2)(3) suy ra:  a = 0,325 ; b = -0,05 ; c = 0,35

→ Sai đề.

Giả sử: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\)

⇒ 24x + 27y = 7,8 (1)

Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

BT e, có: 2x + 3y = 0,8 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,1.24=2,4\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{2,4}{7,8}.100\%\approx30,77\%\\\%m_{Al}\approx69,23\%\end{matrix}\right.\)

b, BTNT Mg và Al, có:

n_MgCl2 = n_Mg = 0,1 (mol)

 n_AlCl3 = n_Al = 0,2 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgCl_2}=\dfrac{0,1.95}{0,1.95+0,2.133,5}.100\%\approx26,24\%\\\%m_{AlCl_3}\approx73,76\%\end{matrix}\right.\)

Bạn tham khảo nhé!

^{trả lời nhanh cho em ạ}

\(a,n_{H_2}=\dfrac{2,576}{22,4}=0,115\left(mol\right)\\ Đặt:n_{Mg}=a\left(mol\right);n_{Al}=b\left(mol\right)\left(a,b>0\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}95a+133,5b=10,475\\a+1,5b=0,115\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,04\\b=0,05\end{matrix}\right.\\ \%m_{Mg}=\dfrac{0,04.24}{0,04.24+0,05.27}.100\approx41,558\%\Rightarrow\%m_{Al}\approx58,442\%\\ b,n_{HCl}=2.n_{H_2}=2.0,115=0,23\left(mol\right)\\ \Rightarrow x=C\%_{ddHCl}=\dfrac{0,23.36,5}{100}.100=8,395\%\)

a, Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\)

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

PTHH:

2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

a---->1,5a--------------------------->1,5a

Mg + H₂SO₄ ---> MgSO₄ + H₂

b------>b----------------------->b

Hệ pt \(\left\{{}\begin{matrix}27a+24b=6,3\\1,5a+b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,15\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{Al}=0,1.27=2,7\left(g\right)\\m_{Mg}=0,15.24=3,6\left(g\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{2,7}{6,3}=42,86\%\\\%m_{Mg}=100\%-42,86\%=57,14\%\end{matrix}\right.\)

b, \(n_{H_2SO_4}=0,1.1,5+0,15=0,3\left(mol\right)\)

\(\rightarrow V_{ddH_2SO_4}=\dfrac{0,3}{0,5}=0,6\left(l\right)=600\left(ml\right)\)

c, đề yêu cầu jv?