\(a)\)Hỗn hợp \(\left\{{}\begin{matrix}Al:a\left(mol\right)\\Fe:b\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow27a+56b=5,5\left(I\right)\)
\(2Al\left(a\right)+6HCl\left(3a\right)\rightarrow2AlCl_3\left(a\right)+3H_2\left(1,5a\right)\)
\(Fe\left(b\right)+2HCl\left(2b\right)\rightarrow FeCl_2\left(b\right)+H_2\left(b\right)\)
\(\sum n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow1,5a+b=0,2\left(II\right)\)
Từ (I) và (II) \(\Rightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,05\left(mol\right)\end{matrix}\right.\)
=> % khối lượng mỗi kim loại
\(b)\)tính thể tích dung dịch HCl tối thiểu cần dùng
Ta có: \(\sum n_{HCl}=3a+2b=3.0,1+2.0,05=0,4\left(mol\right)\)
\(\Rightarrow m_{HCl}=14,6\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{14,6.100}{14,6}=100\left(g\right)\)
\(\Rightarrow V_{ddHCl}=\dfrac{100}{1,08}=92,59\left(ml\right)\)
\(c)\)
\(m dd sau = 5,5+100-0,2.2=105,1(g)\)
Dung dịch sau: \(\left\{{}\begin{matrix}AlCl_3:0,1\left(mol\right)\\FeCl_2:0,05\left(mol\right)\end{matrix}\right.\)
=> C% mỗi muối
