PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

               a_____3a_______a______\(\dfrac{3}{2}\)a   (mol)

            \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

               b_____2b_______b_____b         (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}27a+24b=2,055\\3a+2b=0,205\cdot1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,045\\b=0,035\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=0,045\cdot133,5=6,0075\left(g\right)\\m_{MgCl_2}=0,035\cdot95=3,325\left(g\right)\end{matrix}\right.\)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

=> n_HCl = 0,8 (mol)

Theo ĐLBTKL: m_A,B + m_HCl = m_muối + m_H2

=> m_A,B = 39,4 + 0,4.2 - 0,8.36,5 = 11 (g)

\(n_{HCl}=0,8.0,5=0,4\left(mol\right);n_{H_2SO_4}=0,8.0,75=0,6\left(mol\right)\)

=> \(n_{Cl^-}=0,4\left(mol\right);n_{SO_4^{2-}}=0,6\left(mol\right)\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Bảo toàn nguyên tố H:

\(n_{HCl}.1+n_{H_2SO_4}.2=n_{H_2}.2+n_{H_2O}.2\)

\(\Leftrightarrow0,4.1+0,6.2=0,2.2+n_{H_2O}.2\)

=>\(n_{H_2O}=0,6\left(mol\right)\)

=> \(n_O=0,6\left(mol\right)\)

\(m_{muối}=m_{kimloai}+m_{Cl^-}+m_{SO_4^{2-}}\)

=>\(m_{kimloai}=88,7-35,5.0,4-0,6.96=16,9\left(g\right)\)

=> \(m=m_{kimloai}+m_O=16,9+0,6.16=26,5\left(g\right)\)

n _HCl =  \(0,796.2\) =1,592 (mol) 
n _H2 = \(\dfrac{4,368}{22,4}\)=0,195 (mol) 
Bảo toàn nguyên tố H

=> n _H2O =\(\dfrac{1,592-0,195.2}{2}\) = 0,601 (mol)
=> m _H2O = 10,818 g
Áp dụng định luật bảo toàn khối lượng :

m _{muối khan} = m _{hỗn hợp} + m _axit - m _H2O - m_H2
=> 26,43 + 1,592.36,5 - 10,818 - 0,195.2
= 73,33 g

Chọn C

Đáp án C.

→ n C l - =   2 n H 2   =   0 , 16   → m m u o i   = 4 , 34   +   0 , 16 . 35 , 5   = 10 , 02