Question

Answer 1

a) $n_{Fe} = \dfrac{2,8}{56} = 0,05(mol)$
$Fe + 2HCl \to FeCl_2 + H_2$

Theo PTHH : $n_{HCl} = 2n_{Fe} = 0,1(mol)$
$\Rightarrow V_{dd\ HCl} = \dfrac{0,1}{2} = 0,05(lít)$

b)

$n_{H_2} = n_{Fe} = 0,05(mol) \Rightarrow V_{H_2} = 0,05.22,4 = 1,12(lít)$

c)

$n_{FeCl_2} = n_{Fe} = 0,05(mol)$
$V_{dd\ sau\ pư} = V_{dd\ HCl} = 0,05(lít)$
$C_{M_{FeCl_2}} = \dfrac{0,05}{0,05} = 1M$