Answer to Question #254657 in Physical Chemistry for Sugandhini

Question

Silver metal can be prepared by reducing its nitrate, AgNO3 with zinc according to the following equation: Zn(s) + 2 AgNO3 (aq) → Zn(NO3 )2 (aq) + 2 Ag(s) What is the percent yield of the reaction if 93.5 grams of Ag was obtained from 150.5 grams of AgNO3

Answer 1

93.5 g Ag * (1 mol Ag/107.8682 g Ag) * (2 mol AgNO3/1 mol Zn) * (150.5 g AgNO3/1 mol AgNO3) =260.91

= 150.5/260.91×100

= 57.68%