# Question 8c92c

May 12, 2016

The limiting reactant is "Ca"_10"F"_2("PO"_4)_6. The theoretical yield of ${\text{CaSO}}_{4}$ is 13.5 g. The percent yield of ${\text{CaSO}}_{4}$ is 290 % (pretty good!). The volume of $\text{HF}$ at STP will
be 338 L.

#### Explanation:

Identify the limiting reactant

The balanced chemical equation is

$\text{Ca"_10"F"_2("PO"_4)_6 + "H"_2"SO"_4 → "3Ca"_3("PO"_4)_2 + "CaSO"_4 + "2HF}$

From "Ca"_10"F"_2("PO"_4)_6 :

${\text{Moles of CaSO"_4 = 100 color(red)(cancel(color(black)("g Ca"_10"F"_2("PO"_4)_6))) × (1 color(red)(cancel(color(black)("mol Ca"_10"F"_2("PO"_4)_6))))/(1008.61 color(red)(cancel(color(black)("g Ca"_10"F"_2("PO"_4)_6)))) × ("1 mol CaSO"_4)/(1 color(red)(cancel(color(black)("mol Ca"_10"F"_2("PO"_4)_6)))) = "0.099 15 mol CaSO}}_{4}$

From ${\text{H"_2"SO}}_{4}$:

$\text{Moles of CaSO"_4 = 0.500 color(red)(cancel(color(black)("L H"_2"SO"_4))) × (0.750 color(red)(cancel(color(black)("mol H"_2"SO"_4))))/(1 color(red)(cancel(color(black)("L H"_2"SO"_4)))) × ("1 mol CaSO"_4)/(1 color(red)(cancel(color(black)("mol H"_2"SO"_4)))) = "0.375 mol CaSO} 4$

"Ca"_10"F"_2("PO"_4)_6 gives fewer moles of ${\text{CaSO}}_{4}$, so "Ca"_10"F"_2("PO"_4)_6 is the limiting reactant.

Theoretical yield of ${\text{CaSO}}_{4}$

${\text{Mass of CaSO"_4 = "0.099 15" color(red)(cancel(color(black)("mol CaSO"_4))) × ("136.2 g CaSO"_4)/(1 color(red)(cancel(color(black)("mol CaSO"_4)))) = "13.5 g CaSO}}_{4}$

Percent Yield of ${\text{CaSO}}_{4}$

"% 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 $\text{HF}$

$\text{Moles of HF" = 7500 color(red)(cancel(color(black)("g Ca"_10"F"_2("PO"_4)_6))) × (1color(red)(cancel(color(black)( "mol Ca"_10"F"_2("PO"_4)_6))))/(1008.61 color(red)(cancel(color(black)("g Ca"_10"F"_2("PO"_4)_6)))) × "2 mol HF"/(1 color(red)(cancel(color(black)("mol Ca"_10"F"_2("PO"_4)_6)))) = "14.87 mol HF}$

$P V = n R T$

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"#