Đáp án B

Gọi số mol Mg, Fe, Al là a, b, c

=> 24a + 56b + 27c = 23,8

PTHH: Mg + 2HCl --> MgCl₂ + H₂ 

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

            Fe + 2HCl --> FeCl₂ + H₂

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

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

            c------------------------->1,5c

=> a + b + 1,5c = \(\dfrac{17,92}{22,4}=0,8\left(mol\right)\)

PTHH: Mg + Cl₂ --t^o--> MgCl₂

             a-->a

            2Fe + 3Cl₂ --t^o--> 2FeCl₃

             b--->1,5b

             2Al + 3Cl₂ --t^o--> 2AlCl₃

             c--->1,5c

=> \(a+1,5b+1,5c=\dfrac{20,16}{22,4}=0,9\left(mol\right)\)

=> a = 0,3; b = 0,2; c = 0,2

=> \(\left\{{}\begin{matrix}m_{Mg}=0,3.24=7,2\left(g\right)\\m_{Fe}=0,2.56=11,2\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)

Gọi \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\\n_{Al}=c\left(mol\right)\end{matrix}\right.\) => 65a + 56b + 27c = 10,65 (1)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            Fe + 2HCl --> FeCl₂ + H₂

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

=> \(n_{H_2}=a+b+1,5c=\dfrac{5,04}{22,4}=0,225\left(mol\right)\) (2)

PTHH: Zn + Cl₂ --t^o--> ZnCl₂

            2Fe + 3Cl₂ --t^o--> 2FeCl₃

            2Al + 3Cl₂ --t^o--> 2AlCl₃

=> \(n_{Cl_2}=a+1,5b+1,5c=\dfrac{5,6}{22,4}=0,25\left(mol\right)\) (3)

(1)(2)(3) => \(\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,05\left(mol\right)\\c=0,05\left(mol\right)\end{matrix}\right.\) => \(\left\{{}\begin{matrix}m_{Zn}=0,1.65=6,5\left(g\right)\\m_{Fe}=0,05.56=2,8\left(g\right)\\m_{Al}=0,05.27=1,35\left(g\right)\end{matrix}\right.\)

a) \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{6,5}{10,65}.100\%=61,033\%\\\%m_{Fe}=\dfrac{2,8}{10,65}.100\%=26,291\%\\\%m_{Al}=\dfrac{1,35}{10,65}.100\%=12,676\%\end{matrix}\right.\)

b) n_HCl = 2a + 2b + 3c = 0,45 (mol)

=> m_HCl = 0,45.36,5 = 16,425 (g)

=> \(a\%=C\%=\dfrac{16,425}{200}.100\%=8,2125\%\)

c) m_{dd sau pư} = 10,65 + 200 - 0,225.2 = 210,2 (g)

=> \(\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,1.136}{210,2}.100\%=6,47\%\\C\%_{FeCl_2}=\dfrac{0,05.127}{210,2}.100\%=3,02\%\\C\%_{AlCl_3}=\dfrac{0,05.133,5}{210,2}.100\%=3,176\%\end{matrix}\right.\)

giúp em vs ạ

Đáp án B.

Đáp án A.