$n_{HCl} = 0,2.1 = 0,2(mol)$
$Fe + 2HCl \to FeCl_2 + H_2$
Theo PTHH :
$n_{Fe} = \dfrac{1}{2}n_{HCl} = 0,1(mol)$
$m_{Fe} = 0,1.56 = 5,6(gam)$
Ta có: \(n_{HCl}=0,2.1=0,2\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
____0,1____0,2 (mol)
\(\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
Bạn tham khảo nhé!