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)

Answer 1

Answer – We are given , mass of rock sample = 5.00 g

Moles of HF gas = 6.60*10^-3 moles

Balanced reaction –

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

Now from the above balanced reaction

2 moles of HF = 2 moles of Ca₅(PO₄)₃F(s)

So, 6.60*10^-3 moles of HF = ?

= 6.60*10^-3 moles of Ca₅(PO₄)₃F(s)

So, mas of Ca₅(PO₄)₃F(s) = 6.60*10^-3 moles * 504.30 g/mol

                                          = 3.33 g

So mass percent of Ca₅(PO₄)₃F(s) in rock sample

  = mass of Ca₅(PO₄)₃F(s) / mass of rock sample * 100 %

= 3.33 g / 5.00 g * 100 %

= 66.6 %

So, the percent mass of fluorapatite in the rock sample is 66.6 %