a)
Chất rắn sau phản ứng là Cu
=> mCu = 0,3 (g)
Gọi số mol Al, Fe là x,y (mol)
=> \(m_{Al}+m_{Fe}=2,22\left(g\right)\)
=> 27x + 56y = 2,22 (1)
\(n_{H_2}=\frac{1,344}{22,4}=0,06\left(mol\right)\)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
_______x------->3x------------------>1,5x____(mol)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
_y----->2y------------------->y______(mol)
=> 1,5x + y = 0,06 (2)
(1)(2) => \(\left\{{}\begin{matrix}x=0,02\left(mol\right)\\y=0,03\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%m_{Al}=\frac{0,02.27}{2,52}.100\%=21,43\%\\\%m_{Fe}=\frac{0,03.56}{2,52}.100\%=66,67\%\\\%m_{Cu}=\frac{0,3}{2,52}.100\%=11,9\%\end{matrix}\right.\)
b)
\(n_{HCl}=3x+2y=0,12\left(mol\right)\)
=> m dd HCl = \(\frac{4,38.100}{25}=17,52\left(g\right)\)
=> V dd HCl = \(\frac{17,52}{1,19}=14,72\left(ml\right)\)
