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

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

\(m=27a+56b=19.3\left(g\right)\left(1\right)\)

\(n_{H^+}=0.2\cdot2+0.2\cdot2.25\cdot2=1.3\left(mol\right)\)

\(2Al+6H^+\rightarrow2Al^{3+}+3H_2\)

\(Fe+2H^+\rightarrow Fe^{2+}+H_2\)

\(n_{H^+}=3a+2b=1.3\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.3,b=0.2\)

\(\%Al=\dfrac{0.3\cdot27}{19.3}\cdot100\%=41.96\%\)

\(\%Fe=58.04\%\)

\(b.\)

\(n_{H_2}=\dfrac{1}{2}n_{H^+}=0.65\left(mol\right)\)

Bảo toàn khối lượng : 

\(m_{Muối}=19.3+0.4\cdot36.5+0.45\cdot98-0.65\cdot2=76.7\left(g\right)\)

em cảm ơn ạ!

Gọi số mol Al, Fe là a, b (mol)

=> 27a + 56b = 11,1 (1)

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

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

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

           Fe + 2HCl --> FeCl₂ + H₂

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

=> 1,5a + b = 0,3 (2)

(1)(2) => a = 0,1; b = 0,15

=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1.27}{11,1}.100\%=24,32\%\\\%m_{Fe}=\dfrac{0,15.56}{11,1}.100\%=75,68\end{matrix}\right.\)

a) n Fe = a(mol) ; n Cu = b(mol)

=> 56a + 64b = 4,8(1)

n SO2 = 2,24/22,4 = 0,1(mol)

Bảo toàn e :

3n Fe + 2n Cu = 2n SO2

<=> 3a + 2b = 0,2(2)

Từ (1)(2) suy ra a = b = 0,04

%m Fe = 0,04.56/4,8  .100% = 46,67%

%m Cu = 100% -46,67% = 53,33%

b)

n KOH = 0,15

Ta có :

1 < n KOH / n SO2 = 0,15/0,1 = 1,5 < 2 nên muối sinh ra là Na2SO3(x mol) và NaHSO3(y mol)

2NaOH + SO2 $\to$ Na2SO3 + H2O

NaOH + SO2 $\to$ NaHSO3

Theo PTHH :

n SO2 = a + b = 0,1

n NaOH = 2a + b = 0,15

Suy ra a= 0,05 ; b = 0,05

m muối = 0,05.126 + 0,05.104 = 11,5 gam

Bài 2 :

- Gọi số mol Fe và Cu lần lượt là a, b mol

Ta có : mhh = mFe + mCu = 56a + 64b = 4,8

Bte : 3a + 2b = 0,2 

=> a = b =0,04mol

a, Ta có : mFe =n.M = 2,24g ( 46,7% )

=> %Cu = 53,3%

b, Ta thấy sau phản ứng thu được K+, SO3-2 xmol, HSO3- y mol

BtS : x + y = 0,1 

BTĐT : 2x + y = 0,15 

=> x =y = 0,05 mol

=> mM = mK2SO3 + mKHSO3 = 13,9g

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

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

giup em

1. \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

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

2. Gọi: \(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\) ⇒ 65x + 27y = 9,2 (1)

Theo PT: \(n_{H_2}=n_{Zn}+\dfrac{3}{2}n_{Al}=x+\dfrac{3}{2}y=\dfrac{5,6}{22,4}=0,25\left(mol\right)\left(2\right)\)

Từ (1) và (2) ⇒ x = y = 0,1 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{9,2}.100\%\approx70,65\%\\\%m_{Al}\approx29,35\%\end{matrix}\right.\)

3. Theo PT: \(\left\{{}\begin{matrix}n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ZnSO_4}=0,1.160=16\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\end{matrix}\right.\)

\(n_{HCl}=2.0,4=0,8(mol)\\ n_{Fe}=x(mol);n_{Al}=y(mol)\\ \Rightarrow 56x+27y=11(1)\\ Fe+2HCl\to FeCl_2+H_2\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow 2x+3y=0,8(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)\)

\(a,\Sigma n_{H_2}=x+1,5y=0,4(mol)\\ \Rightarrow V_{H_2}=0,4.22,4=8,96(l)\\ b,m_{Fe}=0,1.56=5,6(g);m_{Al}=0,2.27=5,4(g)\\ c,m_{dd_{HCl}}=400.1,12=448(g)\\ n_{FeCl_2}=0,1(mol);n_{AlCl_3}=0,2(mol)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,1.127}{5,6+448-0,1.2}.100\%=2,8\%\\ C\%_{AlCl_3}=\dfrac{0,2.133,5}{5,4+448-0,3.2}.100\%=5,9\%\)

1. \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

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

2. Gọi: \(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\) ⇒ 65x + 27y = 17,7 (1)

Theo PT: \(n_{H_2}=n_{Zn}+\dfrac{3}{2}n_{Al}=x+\dfrac{3}{2}y=\dfrac{5,6}{22,4}=0,25\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{66}{235}\\y=-\dfrac{29}{1410}\end{matrix}\right.\)

Tới đây thì ra số mol âm, bạn xem lại đề nhé.

Đặt \(\left\{{}\begin{matrix}n_{Al}=x\\n_{Fe}=y\end{matrix}\right.\) ( mol ) \(\rightarrow m_{hh}=27x+56y=5,54\left(g\right)\)  (1)

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

 x                                                   1,5x    ( mol )

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

 y                                           y       ( mol )

\(n_{H_2}=1,5x+y=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)   (1)

\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,06\\y=0,07\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,06.27}{5,54}.100=29,24\%\\\%m_{Fe}=100-29,24=70,76\%\end{matrix}\right.\)