Question #c9ede
1 Answer
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
#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
In other words, for every
#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
#2.84 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.