Đổi: V$H_{2}$= 224ml=0,224l
Cu không phản ứng
Fe + 2HCl -> Fe$Cl_{2}$ + $H_{2}$
0,01 0,02 0,01 0,01 (mol)
n$H_{2}$=$\frac{V}{22,4}$=$\frac{0,224}{22,4}$=0,01(mol)
$m_{Fe}$=n.M=0,01.56=0,56(g)
-> %$m_{Fe}$=$\frac{m_{Fe}}{m_{hh}}$.100=$\frac{0,56}{12}$.100≈4,67%
$m_{Cu}$=$m_{hh}$-$m_{Fe}$=12-0,56=11,44(g) -> %$m_{Cu}$=$\frac{m_{Cu}}{m_{hh}}$.100=$\frac{11,44}{12}$.100≈95,33(%)