Answer to Question #96545 in General Chemistry for Kaylee Williams

Question

A 5.86 g mixture contains both lithium fluoride, LiF, and potassium fluoride , KF. If the mixture contains 3.89 g fluorine, what is the mass of the KF in the mixture?

Answer 1

Let moles of LiF to be x mol, and moles of KF to be y moles.

Then m(LiF) = M*n = 25.94x, m(KF) = M*n = 58.10y

As 1 molecule of LiF contains 1 atom of F , then n(F) in LiF =x mol, and m(F) in LiF = 19x

As 1 molecule of KF contains 1 atom of F, then n(F) in KF = y mol, and m(F) in KF = 19 y.

Solve the system of equations:

$$\begin{alignedat}{2} 25.94&x+ &58.10&y = 5.86 \\ 19&x+&19&y = 3.89 \end{alignedat}$$

x= 0.188, y = 0.0169

n(KF) = y = 0.0169 mol

m(KF) = 0.0171*58.10 = 0.982 g