Respuesta :
Answer: [tex]Li_2O[/tex] is a limiting reactant.
Explanation:
[tex]\text{moles}=\frac{\text{given mass}}{\text{Molecular mass}}[/tex]
[tex]\text{moles of }Li_2O=\frac{65\times 1000g}{29.88gmole^-1 }[/tex]
[tex]=2175.36moles[/tex]
[tex]\text{moles of }H_2O=\frac{80\times 1000g}{18.01gmole^-1}[/tex]
[tex]=4441.97moles[/tex]
As can be seen from the balanced equation 1 mole of [tex]Li_2O[/tex] is required to remove 1 mole of [tex]H_2O[/tex] .
Thus 2175.36 moles of [tex]Li_2O[/tex] will only be able to remove 2175.36 moles of [tex]H_2O[/tex]. Thus [tex]Li_2O[/tex] limit the formation of products and is called as limiting reagent.
(4441.97-2175.36) = 2266.61 moles of water are present in excess and thus water is here present as an excess reagent.
