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

How many grams of fluorine must be reacted with excess lithium iodide to produce 100 grams of lithium fluoride class=

Respuesta :

transitionstate
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

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