Question

For the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

2Li(s) + F2(g) → 2LiF(s)

A) 1.0 g Li; 1.0 g F2

B) 10.5 g Li; 37.2 g F2

C) 2.85×103 g Li; 6.79×103 g F2

the limiting reactants for each

Answer 1

Solution-

First, convert the moles in the equation to grams:
(2 mol of Li)* (6.9410 g of Li/mol) = 13.882 g
1 mol of F2 = 37.9968 g
So when the Li and F are stoichiometrically balanced the ratio of theiar masses is:
13.882 g / 37.9968 g = 0.365347

Now look at each pair:
1.0 g of Li / 1.0 g of F2 = 1
That is greater than the stoichiometric ratio, so Li is in excess and F2 is the limiting reactant

10.5 g of Li/ 37.2 g of F2 = 0.28225
So Li is the limiting reactant.

2.85×10^3 g of Li / 6.79×10^3 g of F2 = 0.41973
So F2 is the limiting reactant.