For the reaction shown, find the limiting reactant for each of the initial quantities of reactants. 2li (s) + f2 (g) → 2lif (s) 1.0 g li; 1.0 g f2 10.5 g li; 37.2 g f2 2.85*103 g li; 6.79*103 g f2

Since, we have the reaction as,

2Li (s) + F2 (g) - - > 2LiF (s)

we are only concerned with the limiting reactants. We calculate for the amount of product that can be produced with the given amount of reactants.

a. 1 g Li (1 mol / 6.941 g of Li) (2 mol LiF/2 mol Li) = 0.144 mol LiF2

1 g F2 (1 mol/38 g) (2 mol LiF2/1 mol F2) = 0.052 mol LiF2

Answer: 1 g of F2

b. 10.5 g Li (1 mol/6.941 g of Li) (2 mol LiF/2 mol Li) = 1.512 mol LiF2

37.2 g F2 (1 mol/38 g) (2 mol LiF2/1 mol F2) = 1.958 mol LiF2

Answer: 10.5 g of Li

c. (2.85 x 10^3 g Li) (1 mol/6.941 g of Li) (2 mol LiF/2 mol Li) = 410.60 mol LiF2

(6.79 x 10^3 g F2) (1 mol/38 g) (2 mol LiF2/1 mol F2) = 357.368 mol of LiF2

Answer: 6.79 x 10^3 g F2