5. Phosphorus fertilizers are derived from phosphate rocks, called fluorapatite, Ca5(PO4)3F(s) . Fluorapatite is insoluble in water, so it must first be dissolved using excess sulfuric acid to form water-soluble calcium dihydrogen phosphate Ca(H2PO4)2. If a powdered 5.00 gram sample of a rock containing fluorapatite is reacted with sulfuric acid and 6.60 x 10-3 moles of HF(g) are released during the process, what is the % mass of fluorapatite in the rock sample?

Question

5.	Phosphorus fertilizers are derived from phosphate rocks, called fluorapatite, Ca5(PO4)3F(s) . Fluorapatite is insoluble in water, so it must first be dissolved using excess sulfuric acid to form water-soluble calcium dihydrogen phosphate Ca(H2PO4)2.  If a powdered 5.00 gram sample of a rock containing fluorapatite is reacted with sulfuric acid and 6.60 x 10-3 moles of HF(g) are released during the process, what is the % mass of fluorapatite in the rock sample?

2 Ca5(PO4)3F(s) + 7 H2SO4(aq)  3 Ca(H2PO4)2 (aq) + 7 CaSO4(aq) + 2 HF (g)

DrBob222 · Answer

mols HF produced = 0.00660
Convert mols HF to mols Ca5(PO4)3F by using the coefficients in the balanced equation.
0.00660 mols HF x (2 mol Ca3(PO4)3F/2 mols HF) = ?
Convert mols Ca5(PO4)3F to grams. g = mols x molar mass and this is the theoretical yield (TY). Assuming 100% efficiency in the process, then % fluorapitite = (mass fluorapitite/mass sample)*100= ?
Post your work if you get stuck.