Question #c9ede

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.