# If 507 g FeCl_2 were used up in the reaction FeCl_2 + 2NaOH -> Fe(OH)_2(s) + 2NaCl, how many grams of Fe(OH)_2 would be formed?

$507 g$ $F e C {l}_{2}$ / $126.751 g$ $F e C {l}_{2}$ = $4 m o l$ $F e C {l}_{2}$
$4 m o l$ $F e C {l}_{2}$ x $1 m o l$ $F e {\left(O H\right)}_{2}$ = $4 m o l$ $F e {\left(O H\right)}_{2}$
$4 m o l$ $F e {\left(O H\right)}_{2}$ x $89.86 g$ $F e {\left(O H\right)}_{2}$ = $359.44 g$ $F e {\left(O H\right)}_{2}$