Question

Mass of 3 MgCl2 - 3(24. 3) (35.5)2 72.9+ 1 Mass of 3 MgCl2 = 144 g 3. What mass of Ca3(PO4)2 is obtained from 15.0 g of Mg3(PO4)2 and 38.0 g of CaCl2? 4. The percent yield of the reaction BiO3 + 3C – 2Bi + 3CO in CO is 90.6%. What is the actual yield of CO if 95.0 g of BiO3 reacts with excess C? lo yield 107

Answer 1

Given mgleoala + 3 Call caz (PO4)2 + 3 Mgu, upon balancing the reaction, we get 1 mole it My, (pot) neacts with 3 moles of call. Also givent mass of mgz (PO4)₂ 2 15.09 mass of cach = 380g moles of Mg (A) = 15 Molar mass of Mg 31 Pod) - 15 moles 262 moles of cada = 38 Molar wass ok cada = 38 111 mole limiting agent = one having less mok storchiometric coefficient M93 (PO4)₂ = 15. u call = 38 262-X1 183 = 0.057 i call = 0.114 Hence; limiting reagent a Mq3 (104) from 1 moles of my ₂ (204), = 1 moke of caz Pop/2 Hence, moles of caz 604), formed = 0.057 moles mass of caz (Po42 Formed = 0.057 X molar mass of (93 (Pol. = 0.057 X 310 - 17.67 mg mass of caz (P04), formed = 17.67g|

Bioz +31 BBi +300 7 gield = go.6% given criven tet mass of Bioz s used = 95.og moles of Bion = 95 Molar mass of Bion - 95 moles 257 from 1 mole of Bioz 3 moles of co formed moles of co formed = 3x95 mole 257 mass of co formed (theoritical) = 3x 95 x 28 = 27.72 g 257 to yield = Actual Theorifrcal Actual = greld & The critical 100 - 90.6 x 27.72 = 25.11g 100
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