A 5.32 g mixture contains both lithium fluoride, LiF, and potassium fluoride, KF. If the mixture contains 3.12 g fluorine, what

Question

kondor19780726 [428]

3 years ago

14

A 5.32 g mixture contains both lithium fluoride, LiF, and potassium fluoride, KF. If the mixture contains 3.12 g fluorine, what

is the mass of the KF in the mixture?

Chemistry

1 answer:

tatiyna · Answer 1 · 2022-07-18T19:27:07+03:00

Answer:

Explanation:

entire mixture = 5.32g

F = 3.12g --> 3.12g*(1mol/18.998g) = 0.1642mols F

Since LiF has 1mol Li to 1 mol F,

0.1642mols F* (1mol Li/1 mol F) = 0.1642mols Li

Convert into grams:

0.1642mols Li * (6.94g/1mol) =1.1397g Li

So add the masses of Li and F together:

3.12g F + 1.1397g Li = 4.2597g LiF

5.32g - 4.2597g LiF = 1.06g KF

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Basically what we did is convert the given mass of F to mols and since we know that there is a 1:1 mole ratio of F to L, the number of mols of F is the number of mols of Li. Change the mols of Li to grams and then add the mass of Li with given mass of F. You now have the mass of LiF. To find mass of KF, subtract mass of LiF from mass of entire mixture.