Question

A 5.00g mixture contains both lithium fluoride, LiF, and potassium fluoride,KF. If the mixture contains 3.10g of fluorine, what is the mass of KF in the mixture

Answer 1

Let the LiF be x then KF will be (5-x)

molar mass of LiF = 26

molar mas of KF = 58

moles of LiF = $\frac{x}{26}$

moles of KF =$\frac{5-x}{58}$

amount of total fluorine present = amount of fluorine present in (LiF +KF)

amount of fluorine present in LiF = $\frac{19}{26}\times\frac{x}{26}$

amount of fluorine present in KF =$\frac{19}{58}\times \frac{5-x}{58}$

total fluorine is equal to 3.10 grams

so lets substitute and get the value of x

$\frac{19}{26}\times\frac{x}{26}+\frac{19}{58}\times \frac{5-x}{58}=3.1$

$19x\times58+19\times26(5-x)=3.1\times58\times26\\$

$1102x+2470-494x=4674.8\\608x=2204.8\\x=3.626\ grams$

mass of KF=$5-x=5-3.626=1.373 grams$