Question

- TUT 73 SHORT ANSWER. For problems where calcul des any equations and/or conversion factors used. No work credit. This 17) (14 points) Ammonia and oxygen react to produce it as and water vapor according to the reaction below In a particular experiment, 3.92g of N, are produced by the reaction of NH with 7 of Os What is the yield of N, in this experiment? NH3(g) + O2(g) N () - H.O() (NOT BALANCED)

Answer 1

page-1 O NH₃(g) + O₂(g) - N₂(8) + H₂0 (2) (Not Balanced] In this reaction, 3.92g of Ny are produced by the reaction of Fg of NH3 and 78 of O₂. What is the % yield of ų is this experiment. on 4 NH3 (3) +30₂ (9) - 2N, (%) + 6H2O (9) Soln given mass of NHg & mass of a > Ag and the actual yield of Wq = 3.92 g now we have to calculate the theoretical yield of N2 (3) - A theoretical yield can tell us how much of a product we will end up with, after completing the reaction. So, 7g ( NHz) x s from stoctometry a from stiochometry 4 mole (NM3 x 17.01 g (NU) 2 mole (N2 (8)) & mole (NU) -= 0.82 mole Ne (8) mass and, s from sto chometry 7910) x s 3 mole (0₂) from stiochiometry x 2 mol (N2) molar ass (15.99 g (o) = & mole (0) - 0.87 mole N₂ (8) now, 0.87 > 0.82 from the (O) NH₂ in our from (NH3) Hence limiting reagent

Page 2 Limiting reagent ! The reagent that will end and used up completely, this limiting the extent of the reaction. Hence by using all the NU (2) we will end up with 0.82 moles of N, (g) - so theoretical yield of a molar mass Nz (8) = 0.82 moles = (0.82 x 28.017419 = 22.97 g 3.92 g And the actual yield = so the % yield - Actual yield theoretical yield 100 3.928 100 = x 22-97 g of yield (N213) = 17%