Answer: 1.35 moles of [tex]Li[/tex] are needed to produce 0.45 mol of [tex]Li_3N[/tex]
Explanation:
The balanced chemical reaction is ;
[tex]6Li(s)+N_2(g)\rightarrow 2Li_3N(s)[/tex]
Thus [tex]Li[/tex] is the limiting reagent as it limits the formation of product and [tex]N_2[/tex] is the excess reagent.
According to stoichiometry :
2 moles of [tex]Li_3N[/tex] require = 6 moles of [tex]Li[/tex]
Thus 0.45 moles of [tex]Li_3N[/tex] will require=[tex]\frac{6}{2}\times 0.45=1.35moles[/tex] of [tex]Li[/tex]
Thus 1.35 moles of [tex]Li[/tex] are needed to produce 0.45 mol of [tex]Li_3N[/tex]
