Answer:
Theoretical yield of silver = 67.97 g Ag
Percent yield of the reaction = 88.3%
Step-by-step explanation:
This reaction forms copper(II) nitrate instead of copper(I) nitrate.
First, we have to write the balanced chemical equation for the reaction of copper (Cu) with silver(I) nitrate (AgNO₃) to give silver (Ag) and copper(I) nitrate (CuNO₃):
Cu(s) + 2AgNO₃(aq) → 2Ag(s) + Cu(NO₃)₂(aq)
From the equation, 1 mol of Cu(s) produces 2 moles of Ag(s). We convert the moles to mass with the molecular weight (MW) of the compounds:
MW(Cu) = 63.5 g/mol
mass of Cu = 1 mol Cu x 63.5 g/mol = 63.5 g Cu
MW(Ag) = 107.9 g/mol
mass of Ag = 2 mol Ag x 107.9 g/mol = 215.8 g Ag
Thus, we have the conversion factor: 215.8 g Ag/63.5 g Cu
- From 20.0 g Cu, the following amount of Ag will be obtained:
20.0 g Cu x 215.8 g Ag/63.5 g Cu = 67.97 g Ag (theoretical amount)
- For an actual amount of 60.0 g of Ag, we calculate the percent yield as follows:
%yield = actual amount/theoretical amount x 100 =
= 60.0 g/67.97 g x 100 = 88.3 %