Question

Phosphoric acid H3PO4 is manufactured by treating mineral phosphates with concentrated sulfuric acid H2SO4. A typical phosphate mineral, Ca5(PO4)3F, reacts with sulfuric acid as shown by the equation:

Ca5(PO4)3F + 5H2SO4  3H3PO4 + 5CaSO4·2H2O + HF
If 5.00 tons of this mineral is treated with 6.00 tons of sulfuric acid to produce 2.70 tons of

phosphoric acid, what is the percent yield of phosphoric acid?

Answer 1

Ca5(PO4)3F + 5H2SO4 +2H2O ========= 3H3PO4 + 5CaSO4*2H2O + HF

The first thing to check is if the equation is balance, and it is. You have 5 atoms of Ca on each side, 3 atoms of P in each side, 1 atom of F in each side, 5 atoms of S in each side, 34 atoms of O in each side, 14 atoms of H in each side.

Inicially we have 5.00tons of Ca5(PO4)3F and we have to change it to mole so we can stablish a relation throught the stequiometrics factors. The way we do the cahnge is with the molecular weight of the cmpound. So we have to change each value of the reactants from tns of grams to mole.

The Molecular weight can be calculate by adding the atomic weight of each atom involving in the compound. For example:

MW: Ca5(PO4)3F= Ca: 5* 40,1g/mol = 200,5g/mol

P: 3 * 31g/mol = 93 g/mol

O: 12 * 16g/mol = 192g/mol

F: 1 * 19g/ml = 19g/mol

MW :Ca5(PO4)3F= 504,5g/mol

Now we calculate with the same procedure the molecular weight of H2SO4, MW= 98g/mol

Now the tons that the problem gave us inicially we make 3 rule

1 mol of Ca5(PO4)3F -----weights-----------504,5g

X ------------------------------------------------------5000000g

X= 9910mol of Ca5(PO4)3F

We do the same with the H2SO4

1mol of H2SO4 -------------weights-------98g

X-------------------------------------------------- 6000000g

X=61224 mole of H2SO4

Now that we have amount of each reactant in moles we stablish the realtion with the stechiometric values from the equation that we had balance before.

1 mole of Ca5(PO4)3F ----- reacts----------- 5 mole of H2SO4

9910mole of Ca5(PO4)3F ---------------------- X

X= 49550 mole of H2SO4

So 9910 mole of Ca5(PO4)3F will react completly with 49550 mole of H2SO4, and so we have an excess of H2SO4 because we calculate that we inicially added 61224mole, so the limitant reactant is the Ca5(PO4)3F because it will ran out first and the the reaction will stop.

Now that we know that Ca5(PO4)3F is the limitant reactant we calculate hw much of the phosphoric acid will form with that value.

1 mol of Ca5(PO4)3F ------------produces ------------- 3mole of H3PO4

9910 mol of Ca5(PO4)3F ------------------------------- X

X= 29730 mol of H3PO4

MW: H3PO4= 98g/mol

1mol of H3PO4 --------weight----------- 98g

29739mol H3PO4 ----------------------X

X= 2913504 g H3PO4

1ton ----------- 1000000g

X --------------- 2913504g

X= 2,91tons

%= 2,70tons /2,91tons *100 = 92,78%

So the percent yild of phosporic acid its 92,78%.