Question

6-85 If 125.5 g of Ca N, were produced from 29.0 g of N, and an excess of Ca according to the reaction 3Ca + N - CazN2 what was the percent yield of Ca N ?

Answer 1

86-85: The chemical reaction is 3ca + N₂ - Ca, N2 Here ca is in excess amount. and Nz is limiting reagent. A limiling reagent (whose amout is lessen than otwa w neagent) determine the amout of product formation . Nz is limiling reagent. 7. yield - Actual yield Theoretical field X 1004. Here actual yield of caz M₂ is 125.58 we have to calculate the theoretical field. from balanced reaction 1 mole of My can form = 1 mole caz M2 i.e. 28 g of My can form = 148 g of caz M2 I g of My can form = 148 g of Cazz ₂,29 g of My can form = 148 x 29 of caz M2 28 = 153.289 i. e. Theoretical we should get 153:28 g of caz M₂

But in expenimently we get only 125.5 g of Ca, M2 They % yield of Ca, M = Actical Yield x 100't. Theoretical yield X (O'L. 125-S 1 153.280 X 100 y. = 81.87.1, ñ 82.4.