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%

\(n_{Cu}=a\left(mol\right),n_{Fe}=b\left(mol\right),n_{Al}=c\left(mol\right)\)

\(m_X=64a+56b+27b=35.7\left(g\right)\left(1\right)\)

\(n_{Cl_2}=\dfrac{21.84}{22.4}=0.975\left(mol\right)\)

\(Cu+Cl_2\underrightarrow{^{^{t^0}}}CuCl_2\)

\(Fe+\dfrac{3}{2}Cl_2\underrightarrow{^{^{t^0}}}FeCl_3\)

\(Al+\dfrac{3}{2}Cl_2\underrightarrow{^{^{t^0}}}AlCl_3\)

\(n_{Cl_2}=a+1.5b+1.5c=0.975\left(mol\right)\left(2\right)\)

\(n_{hh}=ka+kb+kc=0.25\left(mol\right)\)

\(n_{H_2}=kb+k\cdot1.5c=0.2\left(mol\right)\)

\(\Leftrightarrow a-0.25b-0.875c=0\left(3\right)\)

\(\left(1\right),\left(2\right),\left(3\right):a=0.3,b=0.15,c=0.3\)

\(\%Cu=\dfrac{0.3\cdot64}{35.7}\cdot100\%=53.78\%\)

\(\%Fe=\dfrac{0.15\cdot56}{35.7}\cdot100\%=23.52\%\)

\(\text{%Al=22.7%}\)

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)
nCl2nCl2 = 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=0,2.6423,8≈53,78%%Cu=0,2.6423,8≈53,78%

%Fe=0,1.5623,8≈23,53%%Fe=0,1.5623,8≈23,53%

 %_Al ≈ 22,69%

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)
nCl2nCl2 = 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=0,2.6423,8≈53,78%%Cu=0,2.6423,8≈53,78%

%Fe=0,1.5623,8≈23,53%%Fe=0,1.5623,8≈23,53%

 %_Al ≈ 22,69%

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

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

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

Trong \(20,4g\) hỗn hợp có: \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\\n_{Al}=c\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow65a+56b+27c=20,4\left(1\right)\)

\(n_{H_2}=\dfrac{10,08}{22,4}=0,45mol\)

\(BTe:2n_{Zn}+2n_{Fe}+3n_{Al}=2n_{H_2}\)

\(\Rightarrow2a+2b+3c=2\cdot0,45\left(2\right)\)

Trong \(0,2mol\) hhX có \(\left\{{}\begin{matrix}Zn:ka\left(mol\right)\\Fe:kb\left(mol\right)\\Al:kc\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow ka+kb+kc=0,2\)

\(n_{Cl_2}=\dfrac{6,16}{22,4}=0,275mol\)

\(BTe:2n_{Zn}+3n_{Fe}+3n_{Al}=2n_{Cl_2}\)

\(\Rightarrow2ka+3kb+3kc=2\cdot0,275\)

Xét thương:

 \(\dfrac{ka+kb+kc}{2ka+3kb+3kc}=\dfrac{0,2}{2\cdot0,275}\Rightarrow\dfrac{a+b+c}{2a+3b+3c}=\dfrac{4}{11}\)

\(\Rightarrow3a-b-c=0\left(3\right)\)

Từ (1), (2), (3)\(\Rightarrow\left\{{}\begin{matrix}a=0,1mol\\b=0,2mol\\c=0,1mol\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Zn}=6,5g\\m_{Fe}=11,2g\\m_{Al}=2,7g\end{matrix}\right.\)

Đáp án : A

Trong 53,75g X có x mol Sn ; y mol Fe ; z mol Al

=> t(119x + 56y + 27z) = 53,75g

X + Cl₂ -> SnCl₄ ; FeCl₃ ; AlCl₃

⇒ t 4 x + 3 y + 3 z = 2 n C l 2 = 2 , 25   m o l

(Trong 0,4 mol lượng chất gấp t lần)

=> 9(119x + 56y + 27z) = 215(4x + 3y + 3z)

=> 211x – 141y – 402z = 0⁽¹⁾

=> x + y + z = 0,4 mol⁽²⁾

n H 2  = x + y + 1,5z = 31/70 (mol) ⁽³⁾

Từ (1,2,3) => z = 0,0857 mol

=> m_Al = 2,314g