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?
First, we write the equation for the reaction: 2AgNO3 + Cu = Cu(NO3)2 + 2Ag We find the amount of the substance of copper: n(Cu) = m(Cu)/M(Cu) n(Cu) = 5 g / 63.5 g/mol = 0.079 mol n(Ag) = 0.079 mol * 2 = 0,158 mol Now we find the theoretically possible mass (Ag): Theoretical m(Ag) = n(Ag) * M(Ag) M(Ag) = 108 g/mol Theoretical m(Ag) = 180 g/mol * 0.158 mol = 28.44 g Yield = Practical m/ theoretical m * 100%