Answer to Question #241018 in General Chemistry for Lfg

Question

A 5.38 g mixture contains both lithium fluoride, LiF, and potassium fluoride, KF. If the mixture contains 3.15 g fluorine, what is the mass of the KF in the mixture?

Answer 1

Given;

Total weight of mixture (LiF+KF)=5.97gms.(LiF+KF)=5.97gms.

Let, mass of KFKF in the mixture be x gms.xgms.

$\implies$ mass of LiFLiF in mixture =(5.97-x)gms.=(5.97−x)gms.

Note:

Atomic weight of F=19gms.F=19gms.

Atomic weight of Li =7gms.Li=7gms.

Atomic weight of K = 39 gms.K=39gms.

moles=mass/(molecular weight)moles=mass/(molecularweight)

Thus, moles of KF=x/58KF=x/58

and moles of LiF = (5.97-x)/26LiF=(5.97−x)/26

1 mol of KF contains 1 mol of F atoms. Similarly, 1 mol of LiF contains 1 mol of F atoms.

Thus,

moles of F inLiF=

molesofLiF=(5.97−x)/26 ---(2)

From (1) & (2), we

Total moles of Fluorine =(x/58)+((5.97-x)/26)=(x/58)+((5.38−x)/2

Hence, total weight of Fluorine in sample = moles*Atomic weight=moles∗Atomicweig

=((x/58)+((5.38-x)/26))*19gms.=((x/58)+((5.38−x)/26))∗19gm

=3.90 gms.=3.90gms. ---(give

Now, solving the equation for x, we get

26x +(5.38*58)-58x=3.9*58*26/19

32x=346.26-312.04

therefore x=34.22/32

=1.07gms