Question

Question #c9ede

Answer 1

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 `"1 kg" = 10^3` `"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 `"0.07807 kg"` of calcium fluoride that react, you need `"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 `"2.48 kg"` of hydrogen fluoride, so you can say that it consumed

`2.84 color(red)(cancel(color(black)("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.