Respuesta :
For 1 mole of MgCl2, it would require 2 moles of KOH. ( 1 : 2 mole ratio)
Since you have 3 moles of KOH, it is in excess, and MgCl2 is the limiting reactant.
Since you have 3 moles of KOH, it is in excess, and MgCl2 is the limiting reactant.
Answer: [tex]MgCl_2[/tex]
Explanation:
[tex]MgCl_2+2KOH\rightarrow Mg(OH)_2+2KCl[/tex]
According to stoichiometry,
1 mole of [tex]MgCl_2[/tex] reacts with 2 moles of [tex]KOH[/tex]
Thus when 1 mole of [tex]MgCl_2[/tex] is added to 3 moles of [tex]KOH[/tex] , (3-2)moles= 1 mole of [tex]KOH[/tex] will remain unused.
Thus [tex]MgCl_2[/tex] is the limiting reagent as it decides the formation of products.
[tex]KOH[/tex] is excess reagent as it remains unused.
