Question

How many grams of fluorine must be reacted with excess lithium iodide to produce 10.0 grams of lithium fluoride?

Answer 1

Answer:
             7.32 g of F₂

Solution:
              The equation is as follow,

                                   2 LiI  +  F₂    →    2 LiF  +  I₂

According to equation,

           51.88 g (2 mole) of LiF is produced from  =  37.99 g (1 mole) F₂
So,
                          10 g of LiF will be produced by  =  X g of F₂

Solving for X,
                      X  =  (10 g × 37.99 g) ÷ 51.88 g

                      X  =  7.32 g of F₂