$n_{Fe}=\dfrac{16,8}{56}=0,3\ (mol)$
$n_{HCl}=0,2.3,5=0,7\ (mol)$
a, Phương trình hóa học:
$Fe+2HCl\to FeCl_2+H_2$
Nhận thấy: $0,3<\dfrac{0,7}2\to Fe$ hết
b, Theo PT:
$n_{H_2}=n_{Fe}=0,3\ (mol)$
$\to V_{H_2}=0,3.22,4=6,72\ (l)$
c, Theo PT:
$n_{HCl\ pứ}=2n_{Fe}=2.0,3=0,6\ (mol)$
$\to n_{HCl\ dư}=0,7-0,6=0,1\ (mol)$
$n_{FeCl_2}=n_{Fe}=0,3\ (mol)$
$V_{dd\ spứ}=V_{dd\ HCl}=0,2\ (l)$
$\to \left\{\begin{matrix}C_{M}(dd\ HCl)=\dfrac{0,1}{0,2}=0,5\ (M)\\C_{M}(dd\ FeCl_2)=\dfrac{0,3}{0,2}=1,5\ (M)\end{matrix}\right.$
