Question

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?

Answer 1

Just use a little stoichiometry.

`507g` `FeCl_2` / `126.751 g` `FeCl_2` = `4 mol` `FeCl_2`

`4 mol` `FeCl_2` x `1 mol` `Fe(OH)_2` = `4 mol` `Fe(OH)_2`

`4 mol` `Fe(OH)_2` x `89.86 g` `Fe(OH)_2` = `359.44 g` `Fe(OH)_2`