Phosphorus fertilizers are derived from phosphate rocks, called florapatite, Ca5(PO4)3F. 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 posdered 5 gram sample of a rock containing fluorapatite is reacted with sulfuric acid and 6.6*10^-3 moles of HF are released during the process, then what is the mass % of of fluorapatite in the rock sample?
2Ca5(PO4)3F+7H2SO4=3Ca(H2PO4)2+7CaSO4+2HF
2 answers
posdered was supposed to be powdered
Thank you for the equation. See your post above and change the equation I proposed to the proper one. Post your work there if you run into trouble and I can help you through it. For this problem,
mols Ca5(PO4)3F = grams/molar mass = 5/504.3 = approx 0.006 but you need a more accurate answer for this and all of the other calculations I've made. I've rounded here and there also.
Then convert mols Ca5(PO4)3F to mols HF. Since 1 mol apatite produces 1 mol HF. mols fluoroapatite = mols HF.
Then grams HF = mols HF x molar mass HF = ?. This is the theoretical yield (TY). The actual yield (AY) from the problem is 0.0066 g.
Then % yield = (AY/TY)*100 = ?
mols Ca5(PO4)3F = grams/molar mass = 5/504.3 = approx 0.006 but you need a more accurate answer for this and all of the other calculations I've made. I've rounded here and there also.
Then convert mols Ca5(PO4)3F to mols HF. Since 1 mol apatite produces 1 mol HF. mols fluoroapatite = mols HF.
Then grams HF = mols HF x molar mass HF = ?. This is the theoretical yield (TY). The actual yield (AY) from the problem is 0.0066 g.
Then % yield = (AY/TY)*100 = ?
1 answer