Question #3bfae

Start by examining the balanced chemical equation for this reaction

#"Ca"_10"F"_2("PO"_4)_text(6(s]) + color(red)(7)"H"_2"SO"_text(4(aq]) -> color(purple)(2)"HF"_text((g]) + 3"Ca"("H"_2"PO"_4)_text(2(s]) + color(blue)(7)"CaSO"_text(4(s])#

You know that the reaction will always consume #color(red)(7)# moles of sulfuric acid for every mole of fluorapatite that takes part in the reaction, and produce #color(blue)(7)# moles of calcium sulftate, #"CaSO"_4#.

When sulfuric acid is in excess, you can assume that all the moles of fluorapatite will actually take part in the reaction. This means that you can use the #1:color(blue)(7)# mole ratio that exists between fluorapatite and calcium sulfate to determine how many moles of the latter will be produced

#8.60color(red)(cancel(color(black)("moles fluorapatite"))) * (color(blue)(7)color(white)(a)"moles CaSO"_4)/(1color(red)(cancel(color(black)("mole fluorapatite")))) = color(green)(|bar(ul(color(white)(a/a)"60.2 moles CaSO"_4color(white)(a/a)|)))#

For part (b), you need to use the molar mass of hydrogen fluoride to determine how many moles must be produced by the reaction

#200.0color(red)(cancel(color(black)("g"))) * "1 mole HF"/(20.0color(red)(cancel(color(black)("g")))) = "10.0 moles HF"#

This time, fluorapatite is in excess, so you're looking for the mole ratio that exists between sulfuric acid and hydrogen fluoride. Notice that you get #color(purple)(2)# moles of hydrogen fluoride for every #color(red)(7)# *moles8 of sulfuric acid that take part in the reaction.

This means that in order to produce #10.0# moles of hydrogen fluoride, you need the reaction to consume

#10.0color(red)(cancel(color(black)("moles HF"))) * (color(red)(7)color(white)(a)"moles H"_2"SO"_4)/(color(purple)(2)color(red)(cancel(color(black)("moles HF")))) = "35.0 moles H"_2"SO"_4#

Finally, use the fact that a solution's molarity tells you how many moles of solute you get in one liter of solution to determine how many liters of your #"3.00-M"# sulfuric acid solution would contain this many moles.

#35.0color(red)(cancel(color(black)("moles H"_2"SO"_4))) * overbrace("1 L solution"/(3.00color(red)(cancel(color(black)("moles H"_2"SO"_4)))))^(color(brown)("the given 3.00 M")) = color(green)(|bar(ul(color(white)(a/a)"11.7 L"color(white)(a/)|)))#

The answers are rounded to three sig figs.