Gọi số mol của Cu, Fe, Al trong 23,8 gam hhX lần lượt là x, y, z mol

→ m_X = 64x + 56y + 27z = 23,8 (1)
\(n_{Cl_2}\) = x + 1,5y + 1,5z = 0,65 (2)
0,25 mol X + HCl → 0,2 mol H2 nên 0,2.(x + y + z) = 0,25.(y + 1,5z) (3)
Từ (1), (2), (3) => x = 0,2 mol; y = 0,1 mol; z = 0,2 mol 

\(\%_{Cu} = \dfrac{0,2. 64}{23,8} \approx 53,78\%\)

\(\%_{Fe} = \dfrac{0,1 .56}{23,8} \approx 23,53\%\)

 %_Al ≈ 22,69%

Đáp án C

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

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

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

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

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

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

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

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

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

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

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

\(\%Fe=58.04\%\)

\(b.\)

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

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!

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