Question

A 5.00 g piece of copper is placed in a solution of silver nitrate (AgNO3) in which there is excess silver nitrate. The product produced in this reaction is copper nitrate, Cu(NO3)2. The mass of the silver produced is 15.2g. What is the percent yield for this reaction?

Answer 1

The equation for this reaction is:

$$\mathrm{Cu} + 2\mathrm{AgNO_3} \rightarrow 2\mathrm{Ag} + \mathrm{Cu(NO_3)_2}$$

The moles ratio of Cu: Ag is 1:2 (from equation).

Amount of Cu is m/Mw, where Mw is molecular mass,

$$n = \frac{m}{Mw} = \frac{5.00}{64} = 0.077 \text{ mol}$$

So in the end of reaction amount of Ag must be $2 \times 0.077 = 0.154$ mol

$$m = \frac{n \times Mw}{n}$$

Theoretical mass of Ag is $0.154 \times 108 = 16.63$

The yield is theor. m/ real m. $\times 100\% = 15.2 / 16.63 \times 100\% = 91.40\%$