Question

Mass of 150ml, beaker: Mass of beaker, filter paper, and 30.10 copper (11) phosphate after drying: Mass of copper (ll) phosphate produced: (actual yield) Percent yield of copper (11) phosphate: B. DETERMINATION OF THE COMPOSITION OF A MIXTURE OF SO- DIUM PHOSPHATE AND SODIUM CHLORIDE 3.28 of o Mass of mixture: Balanced chemical equation: 2NA PO + 3aul -> Cua (P)+6 Nace 61.52 32 220.65% Mass of CuCl, necessary: (show calculation) 329. Nas fou wel Nesply 13 mol Cucts 120.489 Cucla 1380.12g Najpo, emol Naz PO, Imol Cucla - Mass of CuCl, used: Mass of filter paper: 0.1899 Mass of beaker: 50.8% 51.9g Total mass after drying: Mass of Cu,(PO.), Mass of N, PO in mixture: (show calculation) Percent Na PO, in mixture: 9 | STOICHIOMETRY OF COPPER (1) PHOSPHATE FORMATION REPORT FORM

Answer 1

Date : Page No. (A Mass Mass Moles of of of CulII) che assigned = 10g Cu(I) a weighed = 1.02g Cu(I) h weighed = 1.02 g molar mass of Calta = 7.6 X103 moles = 1.02 g 134.45 g/mol Mass of 1.85 g Sodium Phosphate (Naz PO4) = weighed Naz Pon weighed 1.85 Moles of - 11.3x10² moles rrrrrrrr111111TYY 163.94 Balanced chemical Equation! 3 Cacda + 2Na, Pou Acc. to stoichiometry (az (P0u) + 6 Nace 3 moles of Call reacts with a moles of Naz poy. - 9 mode of clach reacts with 2/ holes of Napoy. > (7.6 X10 3 mole) of Calle reacts with (2*x 7-6x1) = moles of Nazlou = 5.1 X103 moles of Nag Poy is in excess, Cu(II) is limiting As Nag poy reagent

Date : Page No. Theoretical yield of bath Cu (Pows: 3 mole Cuch produce 1 mole of Cuz (PO4)2. = 7:6810-3 mole Cuchy produce = (1 x 7.6 X10 mole - of Cuz (PO4)₂ Theoretical yield = 2.53x108 mole = 2.53x10 3 x molar mass of Cuz (Pou) Therretical yield = 2.53x10 x 380.58 g = 964.14 X10 3g = [0.96 g Given: Mass of filter paper = 0.1829 Mass of beaker = 68.99 Total mass of all 3 = 70.1 g Mass of Cu₂ (104) produced a 70.1-(0.182 +68.9) I actual yield) To 1.0189 17 yield - 1.018_xloo - 106:04:41 0.96 ₃ is quite Wigher due to some error or y. yield impurities.

w Date: Page No. Part -B Mass of mixture = 3.28g 3.289 = (3. (3.28 mole of Ng Pay) 1763.94 cuce necessary acc. to balanced equation given in part A Mass of Chemical = 3 X 3.28 = 0.03 Mole of Cuchy 63.94 2003 x 134.45 g = 4.049 Required CuCl₂ = 4:04g (if only Nag Pou is ) present in mixture A wwUUUUU - Mass of Cup (PO) = 51.9-(50.8+0.182) = 0.98 g = 0.9189 | Mass of Na pou in mixture : Since I mole Cup (pou) is produced from 2 mole of Nag Pou. » 380.50g of Cuz (100) = 2x 163.94 g of 0.918 g of Cuz (PO4)₂ = 2 X1 3.44x0.918 g 380.58 = Naz Pou in 0.79 g of Nag poy. mixture = 0.79 x 100 = 24.1%. % 3.28