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

b) \(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

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

_____0,02<---0,03<---------------------0,03

=> \(\left\{{}\begin{matrix}\%Al=\dfrac{0,02.27}{2,16}.100\%=25\%\\\%Cu=100\%-25\%=75\%\end{matrix}\right.\)

c) m_H2SO4 = 0,03.98 = 2,94 (g)

=> \(C\%\left(H_2SO_4\right)=\dfrac{2,94}{200}.100\%=1,47\%\)

Gọi $n_{Fe} = a(mol), n_{Zn} = b(mol) , n_{Al} = c(mol) \Rightarrow 56a + 65b + 27c = 20,4(1)$

$Fe + H_2SO_4 \to FeSO_4 + H_2$
$Zn + H_2SO_4 \to ZnSO_4 + H_2$
$2Al +3 H_2SO_4 \to Al_2(SO_4)_3 +3 H_2$

Theo PTHH : $n_{H_2} = a + b + 1,5c = \dfrac{10,08}{22,4} = 0,45(mol)(2)$

Mặt khác : 

$2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3$
$Zn + Cl_2 \xrightarrow{t^o} ZnCl_2$
$2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3$

Theo PTHH : $n_{Cl_2} = 1,5n_{Fe} + n_{Zn} + 1,5n_{Al}$

Suy ra : \dfrac{1,5a + b + 1,5c}{a + b + c} = \dfrac{0,275}{0,2}(3)$

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

$\%m_{Fe} = \dfrac{0,2.56}{20,4}.100\% = 54,9\%$

$\%m_{Zn} = \dfrac{0,1.65}{20,4}.100\% = 31,9\%$
$\%m_{Al} = 100\% - 54,9\% - 31,9\% = 13,2\%$

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)\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

Mol:       x                                                     1,5x

PTHH: Mg + H₂SO₄ → MgSO₄ + H₂

Mol:      y                                                 y

Ta có: \(\left\{{}\begin{matrix}27x+24y=5,1\\1,5x+y=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)

\(\%m_{Al}=\dfrac{0,1.27.100\%}{5,1}=52,94\%;\%m_{Mg}=100-52,94=47,06\%\)

b) 

PTHH: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

Mol:      0,1      0,15                  0,05                            

PTHH: Mg + H₂SO₄ → MgSO₄ + H₂

Mol:     0,1       0,1                 0,1

\(m_{ddH_2SO_4}=\dfrac{\left(0,1+0,15\right).98.100}{9,8}=250\left(g\right)\)

m_{dd sau pứ} = 5,1+250-0,15.2 = 254,8(g)

\(C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{0,05.342.100\%}{254,8}=6,71\%\)

\(C\%_{ddMgSO_4}=\dfrac{0,1.120.100\%}{254,8}=4,71\%\)

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

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

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

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

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

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

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

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

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

\(\%Fe=58.04\%\)

\(b.\)

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

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!

PTHH: 

Zn + H₂SO₄ ---> ZnSO₄ + H₂ (1)

2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂ (2)

Ta có: \(n_{H_2}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)

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

a. Theo PT₍₁₎: \(n_{H_2}=n_{Zn}=x\left(mol\right)\)

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

=> \(x+\dfrac{3}{2}y=0,8\) (*)

Theo đề, ta có: 65x + 27y = 3,79 (**)

Từ (*) và (**), ta có HPT:

\(\left\{{}\begin{matrix}x+\dfrac{3}{2}y=0,8\\65x+27y=3,79\end{matrix}\right.\)

(Ra số âm, bn xem lại đề nhé.)

\(a.n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ Đặt:\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ \rightarrow\left\{{}\begin{matrix}27a+24b=5,1\\1,5a+b=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \left\{{}\begin{matrix}\%m_{Al}=\dfrac{27.0,1}{5,1}.100\approx52,941\%\\\%m_{Mg}\approx47,059\%\end{matrix}\right.\)

\(b.m_{ddH_2SO_4}=\dfrac{0,25.98.100}{9,8}=250\left(g\right)\\ m_{ddsau}=m_{Al,Mg}+m_{ddH_2SO_4}-m_{H_2}=5,1+250-0,25.2=254,6\left(g\right)\\ C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{0,05.342}{254,6}.100\approx6,716\%\\ C\%_{ddMgSO_4}=\dfrac{0,1.120}{254,6}.100\approx4,713\%\)

\(n_{H_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

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

\(0.2.....................................0.2\)

\(m_{Fe}=0.2\cdot56=11.2\left(g\right)\) 

\(m_{Fe}=m_{hh}=11.2\left(g\right)\)

=> Đề sai 

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