Question #8c92c

1 Answer
May 12, 2016

Answer:

The limiting reactant is #"Ca"_10"F"_2("PO"_4)_6#. The theoretical yield of #"CaSO"_4# is 13.5 g. The percent yield of #"CaSO"_4# is 290 % (pretty good!). The volume of #"HF"# at STP will
be 338 L.

Explanation:

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"#