a) Đặt \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow27a+56b=11\)  (1)

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

Bảo toàn electron: \(3a+2b=0,4\cdot2=0,8\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2\cdot27}{11}\cdot100\%\approx49,09\%\\\%m_{Fe}=50,91\%\end{matrix}\right.\)

b) Bảo toàn nguyên tố: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{muối}=m_{AlCl_3}+m_{FeCl_2}=0,2\cdot133,5+0,1\cdot127=39,4\left(g\right)\)

c) Bảo toàn electron: \(3\cdot0,2+3\cdot0,1=2n_{SO_2}\)

\(\Rightarrow n_{SO_2}=0,45\left(mol\right)\) \(\Rightarrow V_{SO_2}=0,45\cdot22,4=10,08\left(l\right)\)

a) Gọi nAl = x, nFe = y

Có 27x + 56y = 11 (1)

Bảo toàn e

3x + 2y = 2.0,4 (2)

Từ 1 và 2 => x = 0,2, y = 0,1

\(\%mAl=\dfrac{0,2.27}{11}.100\%=49,09\%\)

\(\%mFe=100-49,09=50,91\%\)

b) BTKL: 

m muối = mkim loại + mHCl - mH2

= 11 + 0,4.2.36,5 - 0,4.2 = 39,4g

c) 

Bảo toàn e

Al => Al⁺³ + 3e                                  S⁺⁶ + 2e => S⁺⁴

0,2               0,6                                             2x          x

Fe => Fe⁺³ + 3e

0,1                  0,3

=> 2x = 0,6 + 0,3 => x = 0,45 mol

=> VSO₂ = 0,45.22,4 = 10,08 lít

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

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

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

Gọi \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

⇒ 27x + 56y = 11 (1)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{3}{2}x+y\left(mol\right)\)

\(\Rightarrow\dfrac{3}{2}x+y=0,4\left(2\right)\)

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

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{11}.100\%\approx49,1\%\\\%m_{Fe}\approx50,9\%\end{matrix}\right.\)

Bạn tham khảo nhé!

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

\(m_{hh}=27a+56b=11\left(g\right)\left(1\right)\)

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

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

\(n_{H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)

\(\Rightarrow1.5a+b=0.4\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.2,b=0.1\)

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

\(\%Al=\dfrac{5.4}{11}\cdot100\%=49.09\%\)

\(\%Fe=50.91\%\)

Ta có n_H2 = \(\dfrac{8,96}{22,4}\) = 0,4 ( mol )

2Al + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂

x...........3x...........x..............1,5x

Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂

y..........2y.............y.........y

=> \(\left\{{}\begin{matrix}27x+56y=11\\1,5x+y=0,4\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

=> m_Al = 0,2 . 27 = 5,4 ( gam )

=> m_Fe = 11 - 5,4 = 5,6 ( gam )

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

Gọi a,b lần lượt là số mol của Al và Fe

pt1: 2Al + 6HCl -> 2AlCl₃ + 3H₂

cứ:; 2..............6................2...........3(mol)

vậy: a-------->3a----------->a------->1,5a(mol)

pt2: Fe +2HCl -> FeCl₂ +H₂

cứ:; 1...........2...........1............1 (mol)

vậy: b------->2b------>b-------->b (mol)

Từ 2 pt và đề ta có hệ phương trình:

1,5a+b=0,4

27a+56b=11

=> a=0,2 (mol) ,b=0,1(mol)

=>m_Al=n.M=0,2.27=5,4(g)

=> m_Fe=m_hh - m_Al=11-5,4=5,6(g)

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

Gọi x; y lần lượt là số mol của Al ; Fe ta có :

Pt:

\(2Al+6HCl\Rightarrow2AlCl_3+3H_2\)

2x.............................................3x

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

y...........................................y

Ta có hệ phương trình:

\(\left\{{}\begin{matrix}27x+56y=11\\3x+y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)

\(m_{AL}=0,1\times27=2,7\left(g\right)\)

\(m_{Fe}=0,1\times56=5,6\left(g\right)\)

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

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

Ta có: \(m_{H_2}=\dfrac{8,96}{22,4}\cdot2=0,8\left(g\right)< m_{hh}=11\left(g\right)\)

\(\Rightarrow\) Sau p/ứ dd tăng \(11-0,8=10,2\left(g\right)\)

a) 2Al + 6HCl → 2AlCl₃ + 3H₂ (1)

Fe + 2HCl → FeCl₂ + H₂ (2)

b) \(n_{H_2}=\frac{8,96}{22,4}=0,4\left(mol\right)\)

Gọi x,y lần lượt là số mol của Al và Fe

Ta có: \(27x+56y=11\) (*)

Theo pT1: \(n_{H_2}=\frac{3}{2}n_{Al}=1,5x\left(mol\right)\)

Theo Pt2: \(n_{H_2}=n_{Fe}=y\left(mol\right)\)

ta có: \(1,5x+y=0,4\) (**)

Từ (*)(**) ta có: \(\left\{{}\begin{matrix}27x+56y=11\\1,5x+y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

Vậy \(n_{Al}=0,2\left(mol\right)\Rightarrow m_{Al}=0,2\times27=5,4\left(g\right)\)

\(n_{Fe}=0,1\left(mol\right)\Rightarrow m_{Fe}=0,1\times56=5,6\left(g\right)\)

\(\%m_{Al}=\frac{5,4}{11}\times100\%=49,09\%\)

\(\%m_{Fe}=\frac{5,6}{11}\times100\%=50,91\%\)

c) Theo Pt1: \(n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=0,2\times133,5=26,7\left(g\right)\)

Theo pt2: \(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\)

\(\Rightarrow m_{FeCl_2}=0,1\times127=12,7\left(g\right)\)

\(\Rightarrow m_{muối}=26,7+12,7=39,4\left(g\right)\)

\(n_{Al}=x;n_{Fe}=y\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow hpt:\left\{{}\begin{matrix}27x+56y=11\\22,4\left(\frac{3x}{2}+y\right)=8,96\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n_{Al}=0,2\left(mol\right)\rightarrow\%m_{Al}=\frac{27.0,2}{11}.100\%=49,09\left(\%\right)\\n_{Fe}=0,1\left(mol\right)\rightarrow\%m_{Fe}=100-49,09=50,91\left(\%\right)\end{matrix}\right.\\ m_M=m_{AlCl_3}+m_{FeCl_2}=0,2.133,5+0,1.127=39,4\left(g\right)\)

a)

Gọi số mol Al, Fe là a, b

=> 27a + 56b = 8,3

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

            a----->3a------->a------>1,5a

            Fe + 2HCl --> FeCl₂ + H₂

            b------>2b------>b----->b

=> \(1,5a+b=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

=> a = 0,1; b = 0,1

=> \(\left\{{}\begin{matrix}\%Al=\dfrac{0,1.27}{8,3}.100\%=32,53\%\\\%Fe=\dfrac{0,1.56}{8,3}.100\%=67,47\%\end{matrix}\right.\)

b) 

\(\left\{{}\begin{matrix}m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\\m_{FeCl_2}=0,1.127=12,7\left(g\right)\end{matrix}\right.\)

=> m_muối = 13,35 + 12,7 = 26,05(g)

c) 

n_HCl = 3a + 2b = 0,5(mol)

=> \(V_{ddHCl\left(PTHH\right)}=\dfrac{0,5}{2}=0,25\left(l\right)\)

=> V_{dd HCl(thực tế)} = \(\dfrac{0,25.110}{100}=0,275\left(l\right)\)

d)

PTHH: 2FeCl₂ + Cl₂ --> 2FeCl₃

              0,1----------------->0,1

=> \(\left\{{}\begin{matrix}m_{FeCl_3}=0,1.162,5=16,25\left(g\right)\\m_{AlCl_3}=13,35\left(g\right)\end{matrix}\right.\)

=> m_muối = 16,25 + 13,35 = 29,6(g)