a, Ta có: 27n_Al + 56n_Fe = 0,83 (1)

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

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

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow n_{Al}=n_{Fe}=0,01\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,01.27}{0,83}.100\%\approx32,53\%\\\%m_{Fe}\approx67,47\%\end{matrix}\right.\)

b, n_H2SO4 = n_H2 = 0,025 (mol)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,025.98}{20\%}=12,25\left(g\right)\)

\(n_{H_2SO_4}=0,2.2=0,4\left(mol\right)\)

Fe + H2SO4 → FeSO4 + H2 

2Al + 3H2SO4 → Al2(SO4)3 + 3H2

Gọi x,y lần lượt là số mol Fe, Al

\(\left\{{}\begin{matrix}56x+27y=11\\x+\dfrac{3}{2}y=0,4\end{matrix}\right.\)

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

=>\(\%m_{Fe}=\dfrac{0,1.56}{11}.100=50,91\%\)

=> %m _Al = 100 - 50,91 =49,09 %

b)Theo PT:  \(n_{H_2}=n_{H_2SO_4}=0,4\left(mol\right)\)

=> \(V_{H_2}=0,4.22,4=8,96\left(l\right)\)

c) \(CM_{FeSO_4}=\dfrac{0,1}{0,2}=0,5M\)

\(CM_{Al_2\left(SO_4\right)_3}=\dfrac{\dfrac{0,2}{2}}{0,2}=0,5M\)

Đặt nAl=a(mol); nFe=b(mol) (a,b>0)

Ta có: nH2=8,96/22,4=0,4(mol)

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

a_________3a____a____1,5a(mol)

Fe +2 HCl -> FeCl2 + H2

b__2b____b____b(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}27a+56b=16,7\\1,5a+b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,25\end{matrix}\right.\)

=> mAl= 0,1.27=2,7(g) =>%mAl= (2,7/16,7).100=16,17%

=> CHỌN B

TN1: Gọi \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\\n_{Cu}=c\left(mol\right)\end{matrix}\right.\)

=> 65a + 56b + 64c = 37 (1)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

             a---------------------->a

            Fe + 2HCl --> FeCl₂ + H₂

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

=> \(a+b=\dfrac{8,96}{22,4}=0,4\) (2)

TN2: Gọi \(\left\{{}\begin{matrix}n_{Zn}=ak\left(mol\right)\\n_{Fe}=bk\left(mol\right)\\n_{Cu}=ck\left(mol\right)\end{matrix}\right.\)

=> ak + bk + ck = 0,15 (3)

\(n_{Cl_2}=\dfrac{3,92}{22,4}=0,175\)

PTHH: Zn + Cl₂ --t^o--> ZnCl₂

           ak-->ak

           2Fe + 3Cl₂ --t^o--> 2FeCl₃

            bk--->1,5bk

            Cu + Cl₂ --t^o--> CuCl₂

             ck-->ck

=> ak + 1,5bk + ck = 0,175 (4)

(1)(2)(3)(4) => \(\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,2\left(mol\right)\\c=0,2\left(mol\right)\\k=0,25\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,2.65}{37}.100\%=35,135\%\\\%m_{Fe}=\dfrac{0,2.56}{37}.100\%=30,27\%\\\%m_{Cu}=\dfrac{0,2.64}{37}.100\%=34,595\%\end{matrix}\right.\)

\(a,n_{Al}=x(mol);n_{Fe}=y(mol)\\ \Rightarrow 27x+56y=11(1)\\ n_{H_2}=\dfrac{8,96}{22,4}=0,4(mol)\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow 1,5x+y=0,4(2)\\ (1)(2)\Rightarrow x=0,2(mol);y=0,1(mol)\\ \Rightarrow \begin{cases} \%_{Al}=\dfrac{0,2.27}{11}.100\%=49,09\%\\ \%_{Fe}=100\%-49,09\%=50,91\% \end{cases}\\ b,\Sigma n_{HCl}=3x+2y=0,8(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,8}{2}=0,4(l)\)

\(\left\{{}\begin{matrix}Fe\\Al\end{matrix}\right.+HCl->\left\{{}\begin{matrix}FeCl2\\AlCl3\end{matrix}\right.+H2\)

Ta có số mol Fe là x , Al là y (mol)

\(\left\{{}\begin{matrix}56x+27y=11\\127x+133,5y=39,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%mFe=\dfrac{0,1.56}{11}=50,9\%\\\%mAl=\dfrac{0,2.27}{11}=49,09\%\end{matrix}\right.\)

Bảo toàn e :

\(2.nH2=2.nFe+3.nAl\Rightarrow nH2=0,4\left(mol\right)\)

\(V=0,4.22,4=8,96\left(l\right)\)

\(nFe=nFeCl2=0,1\left(mol\right)\)

\(nAl=nAlCl3=0,2\left(mol\right)\)

\(\Rightarrow nHCl\left(pứ\right)=2.0,1+3.0,2=0,8\left(mol\right)\)

\(Cm=\dfrac{n}{V}=\dfrac{0,8}{0,2}=4\left(M\right)\)

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

              a_______a_______a_____a    (mol)

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

                2b______3b__________b_____3b    (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}56a+27\cdot2b=11\\a+3b=0,2\cdot2=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1\cdot56}{11}\cdot100\%\approx50,91\%\\\%m_{Al}=49,09\%\end{matrix}\right.\)

Theo các PTHH: \(\left\{{}\begin{matrix}n_{H_2}=0,4\left(mol\right)\\n_{FeSO_4}=0,1\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,4\cdot22,4=8,96\left(l\right)\\C_{M_{FeSO_4}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\\C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\end{matrix}\right.\) 

a) nH2SO4=0,4(mol)

Đặt: nFe=x(mol); nAl=y(mol) (x,y>0)

PTHH: Fe + H2SO4 -> FeSO4 + H2

x________x______x______x(mol)

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

y____1,5y_______0,5y_______1,5y(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}56x+27y=11\\x+1,5y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

=> mFe=0,1.56=5,6(g)

=>%mFe=(5,6/11).100=50,909%

=>%mAl= 49,091%

b) V(H2,đktc)=0,4.22,4=8,96(l)

c) nAl2(SO4)3= 0,5y=0,5.0,2=0,1(mol)

nFeSO4=x=0,1(mol)

Vddsau=VddH2SO4=0,2(l)

=>CMddAl2(SO4)3= 0,1/0,2=0,5(M)

CMddFeSO4=0,1/0,2=0,5(M)

a) Gọi số mol Al, Fe là a, b (mol)

=> 27a + 56b = 13,9 (1)

\(n_{H_2}=\dfrac{7,84}{22,4}=0,35\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

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

             Fe + 2HCl --> FeCl₂ + H₂

             b-->2b---------------->b

=> 1,5a + b = 0,35 (2)

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

n_HCl = 3a + 2b = 0,7 (mol)

=> m_HCl = 0,7.36,5 = 25,55(g)

b) 

m_Al = 0,1.27 = 2,7 (g)

m_Fe = 0,2.56 = 11,2 (g)