Đáp án là D. 38,55%.

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)

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

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

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

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

Sửa: $V_{H_2}=7,168(l)$

$a\bigg)$

Đặt $n_{Mg}=x;n_{Fe}=y;n_{Al}=z$

$\to 24x+56y+27z=9,52(1)$

$n_{H_2}=\dfrac{7,168}{22,4}=0,32(mol)$

$n_{Cl_2}=\dfrac{8,064}{22,4}=0,36(mol)$

BTe: $x+y+1,5z=n_{H_2}=0,32(2)$

BTe: $x+1,5y+1,5z=n_{Cl_2}=0,36(3)$

Từ $(1)(2)(3)\to x=0,12(mol);y=0,08(mol);z=0,08(mol)$

$\to \begin{cases} \%m_{Mg}=\dfrac{0,12.24}{9,52}.100\%=30,25\%\\ \%m_{Fe}=\dfrac{0,08.56}{9,52}.100\%=47,06\%\\ \%m_{Al}=100-47,06-30,25=22,69\% \end{cases}$

$b\bigg)$

Bảo toàn H: $n_{HCl}=2n_{H_2}=0,64(mol)$

$\to C_{M_{HCl}}=\dfrac{0,64}{0,2}=3,2M$

$\to a=3,2$

$c\bigg)$

Dung dịch sau gồm $MgCl_2,FeCl_2,AlCl_3$

Bảo toàn $Mg,Al,Fe:n_{MgCl_2}=0,12(mol);n_{AlCl_3}=n_{FeCl_2}=0,08(mol)$

$\to C_{M_{MgCl_2}}=\dfrac{0,12}{0,2}=0,6M$

$\to C_{M_{AlCl_3}}=C_{M_{FeCl_2}}=\dfrac{0,08}{0,2}=0,4M$

$a\bigg)$

Đặt $n_{Mg}=x;n_{Fe}=y;n_{Al}=z$

$\to 24x+56y+27z=9,52(1)$

$n_{H_2}=\dfrac{14,336}{22,4}=0,64(mol)$

$n_{Cl_2}=\dfrac{8,064}{22,4}=0,36(mol)$

BTe: $x+y+1,5z=n_{H_2}=0,64(2)$

BTe: $x+1,5y+1,5z=n_{Cl_2}=0,36(3)$

Từ $(1)(2)(3)\to$ nghiệm âm, xem lại đề

giúp em vs

Fe+2HCl->FeCl₂+H₂

x---2x-----------x

Mg+2HCl->MgCl₂+H₂

y------2y-----------y

Ta có :

\(\left\{{}\begin{matrix}56x+24y=24\\x+y=\dfrac{13,44}{22,4}\end{matrix}\right.\)

=>x=0,3 mol, y=0,3 mol

=>%m Fe=\(\dfrac{0,3.56}{24}.100\)=70%

=>%m Mg=100-70=30%

=>VHCl=\(\dfrac{0,3.2+0,3.2}{2}\)=0,6l=600ml

b)

 XCl₂+2AgNO₃->2AgCl+X(NO₃)₂

0,6--------------------1,2mol

=>m AgCl=1,2.143,5=172,2g

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

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