Answer: The number of moles of [tex]Fe(OH)_2[/tex] produced are, 3.60 moles.
Explanation :
First we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]Fe(s)+2NiO(OH)(s)+2H_2O(l)\rightarrow Fe(OH)_2(s)+2Ni(OH)_2(aq)[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]Fe[/tex] react with 2 mole of [tex]NiO(OH)[/tex]
So, 3.60 moles of [tex]Fe[/tex] react with [tex]3.60\times 2=7.2[/tex] moles of [tex]NiO(OH)[/tex]
From this we conclude that, [tex]NiO(OH)[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Fe[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]Fe(OH)_2[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]Fe[/tex] react to give 1 mole of [tex]Fe(OH)_2[/tex]
So, 3.60 mole of [tex]Fe[/tex] react to give 3.60 mole of [tex]Fe(OH)_2[/tex]
Therefore, the number of moles of [tex]Fe(OH)_2[/tex] produced are, 3.60 moles.
