Answer to Question #254537 in Chemistry for Katerina

Question

Silver metal can be prepared by reducing its nitrate, AgNO 3 with zinc according to the

following equation:

Zn(s) + 2 AgNO 3 (aq) → Zn(NO 3 ) 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 AgNO 3 ?

Answer 1

According to the equation, n (AgNO₃) = n (Ag).

n = m / M

M (Ag) = 107.9 g / mol

M (AgNO₃) = 169.9 g / mol

n (AgNO₃) = 150.5 / 169.9 = 0.9 mol

n (AgNO₃) = n (Ag) = 0.9 mol

m (Ag) = n x M = 0.9 x 107.9 = 97.1 g

% (Ag) = (93.5 / 97.1) x 100 = 96.3%