Question

Assuming an efficiency of 26.20%, calculate the actual yield of magnesium nitrate formed from 122.7 g of magnesium and excess copper(II) nitrate. Mg +Cu(NO,)2 Mg(NO,)2 +Cu actual yield:

Answer 1

Ane-i Given Reaction Mg + CULHO/₂ - Mg(NO3)2 + cu Imole of mg reacts with Imole of cu(NO3)2 to give Imole of Mg(NO3)2 and Imole of cu according to balanced reaction Amount of 'Mg = m = 122-79 Molar mass of 'Mo' = 24.305U No of moles of 5.048 mole Mg 24.3050 122.79 CULN03/2 is in excess Mg is limiting Reactant No of moles of Mg (19/ = 5.048 mol . Molar mass of Amount of Mg(NO3)2 formed by a 5.048mole Mg Mg (NO3)2 = 148.3g/mol formed Theoretically - 5.048 mole x 148.3g/mole

Amount of Mg(NO) formed Theoretically = 748-629 Efficiency of - Efficiency % = - Reaction = 26% Amount of Mg(NO3), formed Actually x 100 - Amant of Mg(NO3l2 formed Theoretically 26 - xx100 748.629 e = 194.64g! Answer le = Actual yield of Mg CN 032 formed =194.6g. Answer