Question

9.- One night you and your friend decided to help in agriculture. The farmer you are helping reacted 90 kg of Calcium fluoride phosphate (Cas(PO4)3F) and 30L of sulfuric acid(H2SO4), so far he has collected 5kg of hydrofluoric acid (HF) and the farmer wants to know the % yield of the HF to estimate the time needed to complete the reaction. Balance the equation, calculate the molecular weight and mols of Cas(PO4)3F and H2SO4 (Density of Sulfuric acid = 1.84 g/cm), Calculate the limiting reagent, the percent yield of HF and the number of molecules of so far. If the density of HF is 1.15 g/cm and the temperature of the 10 L vessel HF is contained is 25 C what is the pressure of the contained in which HF is collected? Cas(PO4)3F (s) + H2SO4 (aq) → Ca(H2PO4)2 (aq) + Caso. (aq) + HF (g)

Answer 1

The balanced equation will be... 26a5 (POA) zf + 74₂ 50 36a (H2PO4)2 + 7la504 + 2HF Molecular weight of cas (PO )3f = 140x5) + 3 (31+69) + 19] g/mol = 504 g/mole Holecular weight of H₂SO4 - = (1+2)+ 3.2+(16x9) g/mol = 98 g/mol Molecular weight of itf = (1+19) gimoy = 200mmor 60 Again, density of H₂SO4 = 1184q1ce .: 1.84 g Hason is present in Bicc 98 g' . I. 1 1*98 19 184 = 53. 2bec 0.05326 L

so, from the balanced eam, we can say that (2x504)g = 1008 g of Cas (004), f will react with H4X105 3 26) L= 0.37282L H2504 and will produce (2002) = 409HF here, 90kg = 90000 g cas(POA) af istaken i.e. 190000 +504) = 178:57 mole which will recact with 1178.57X7 - ) = 625 mole of H₂ soa nie. 33.28 L of H2SO4 : But only 3ol of H₂SO4 is given so, the total car (PO4)₂f lan't be reeacted: 30L of H2SO4 nie. (30 - 05326) mode = 563.274 mole - H₂SO will react with (5.6 3 274x2) - 160.93 mole cas (POA) of jie. (160.93 x 504) g = 81108.72

So, after the reaction total Hy sou will be consumed. so, - H₂SO is limiting reagent and 81108.72 g of car (PO4). 3. f will react and the remaining ine. 90000 - 81108.72) g = 8891.289 or. 8.89 kg of cas (POA), of will термохи л гоосоо 7 from the reaction of 563.27 4 mole of t₂ soa and 160.93 mole of Car (POA) 3f, the product is 160.93 mole of HF ive. /20 x 160193) g = 3218.69 : 5000-321816 - 3218261100) .255.347.

A Temperature (T): 250 = (273+25)K 290K volume (u) = 10L ug kи 0$, Pt Pressure S PV=nRT vyvolume ny no. of mole . RT of gas on. pv=un Tant emparature Ra Universal Gas or PM = m RT - consta my mass of species on PM = dRT May Moleculen weight of species] al density px 20 =1157.813 14x107x298 on pe 5497 632'5 Pa