Identify the limiting reactant
The balanced chemical equation is
#"Ca"_10"F"_2("PO"_4)_6 + "H"_2"SO"_4 → "3Ca"_3("PO"_4)_2 + "CaSO"_4 + "2HF"#
From #"Ca"_10"F"_2("PO"_4)_6 #:
#"Moles of CaSO"_4 = 100 color(red)(cancel(color(black)("g Ca"_10"F"_2("PO"_4)_6))) × (1 color(red)(cancel(color(black)("mol Ca"_10"F"_2("PO"_4)_6))))/(1008.61 color(red)(cancel(color(black)("g Ca"_10"F"_2("PO"_4)_6)))) × ("1 mol CaSO"_4)/(1 color(red)(cancel(color(black)("mol Ca"_10"F"_2("PO"_4)_6)))) = "0.099 15 mol CaSO"_4#
From #"H"_2"SO"_4#:
#"Moles of CaSO"_4 = 0.500 color(red)(cancel(color(black)("L H"_2"SO"_4))) × (0.750 color(red)(cancel(color(black)("mol H"_2"SO"_4))))/(1 color(red)(cancel(color(black)("L H"_2"SO"_4)))) × ("1 mol CaSO"_4)/(1 color(red)(cancel(color(black)("mol H"_2"SO"_4)))) = "0.375 mol CaSO"4#
#"Ca"_10"F"_2("PO"_4)_6# gives fewer moles of #"CaSO"_4#, so #"Ca"_10"F"_2("PO"_4)_6# is the limiting reactant.
Theoretical yield of #"CaSO"_4#
#"Mass of CaSO"_4 = "0.099 15" color(red)(cancel(color(black)("mol CaSO"_4))) × ("136.2 g CaSO"_4)/(1 color(red)(cancel(color(black)("mol CaSO"_4)))) = "13.5 g CaSO"_4#
Percent Yield of #"CaSO"_4#
#"% yield" = "actual yield"/"theoretical yield" × 100 % = (39.2 color(red)(cancel(color(black)("g"))))/(13.5 color(red)(cancel(color(black)("g")))) × 100 % = 290 %#
Yield of #"HF"#
#"Moles of HF" = 7500 color(red)(cancel(color(black)("g Ca"_10"F"_2("PO"_4)_6))) × (1color(red)(cancel(color(black)( "mol Ca"_10"F"_2("PO"_4)_6))))/(1008.61 color(red)(cancel(color(black)("g Ca"_10"F"_2("PO"_4)_6)))) × "2 mol HF"/(1 color(red)(cancel(color(black)("mol Ca"_10"F"_2("PO"_4)_6)))) = "14.87 mol HF"#
#PV = nRT#
#V = (nRT)/P = (14.87 color(red)(cancel(color(black)("mol"))) × "0.083 14" color(red)(cancel(color(black)("bar")))·"L"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar")))) = "338 L"#
