Question

[Ha Nguyen Thi]

Giúp mình câu 3 với

Answer 1

[Kudo Shinichi]

Câu 3:

$a) \rm n_{Fe} = \dfrac{84}{56} = 1,5 (mol)$
PTHH:
$\rm Fe + 2HCl \rightarrow FeCl_2 + H_2$
Theo PT: $\rm n_{FeCl_2} = n_{Fe} = 1,5 (mol)$
$\rm m_{FeCl_2} = 1,5.127 = 190,5 (g)$
$\rm b)$ Theo PT: $\rm n_{KOH} = 2n_{FeCl_2} = 3 (mol)$
$\rm \Rightarrow C_{M(KOH)} = \dfrac{3}{\dfrac{200}{1000}} = 15M$

Answer 2

[Kudo Shinichi]

Quên PTHH ._. 

$\rm FeCl_2 + 2KOH \rightarrow Fe(OH)_2 \downarrow + 2KCl$