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

Chemistry

1 answer:

rewona [7]3 years ago

8 0

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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