# Question c9ede

Apr 22, 2017

Here's what I got.

#### Explanation:

Start by looking up the molar masses of calcium fluoride, sulfuric acid, and hydrogen fluoride

M_ ("M CaF"_2) = "78.07 g mol"^(-1)

M_ ("H"_ 2"SO"_ 4) = "98.08 g mol"^(-1)

M_ ("HF") = "20.01 g mol"^(-1)

Convert these values from grams per mole to kilograms per mole -- remember that $\text{1 kg} = {10}^{3}$ $\text{g}$

M_ ("M CaF"_2) = "0.07807 kg mol"^(-1)

M_ ("H"_ 2"SO"_ 4) = "0.09808 kg mol"^(-1)

M_ ("HF") = "0.02001 kg mol"^(-1)

Now, notice that for every $1$ mole of calcium fluoride that takes part in the reaction, the reaction consumes $1$ mole of sulfuric acid and produces $2$ moles of hydrogen fluoride.

In other words, for every $\text{0.07807 kg}$ of calcium fluoride that react, you need $\text{0.09808 kg}$ of sulfuric acid and you get

2 color(red)(cancel(color(black)("moles HF"))) * "0.02001 kg"/(1color(red)(cancel(color(black)("mole HF")))) = "0.04002 kg"

of hydrogen fluoride.

In your case, you know that the reaction produced $\text{2.48 kg}$ of hydrogen fluoride, so you can say that it consumed

$2.84 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{kg HF"))) * "0.07807 kg CaF"_2/(0.04002color(red)(cancel(color(black)("kg HF")))) = color(darkgreen)(ul(color(black)("5.5 kg CaF}}_{2}}}}$

This means that the percent purity of the sample was

"% CaF"_2 = (5.5 color(red)(cancel(color(black)("kg"))))/(6.0color(red)(cancel(color(black)("kg")))) xx 100% = color(darkgreen)(ul(color(black)(92%)))#

The values are rounded to two sig figs, the number of sig figs you have for the mass of calcium fluoride and of sulfuric acid.