When copper wire is placed into a silver(I) nitrate solution, silver crystals and copper(I) nitrate solution form. If a 20.0g sample of copper is used, determine the theoretical yield of silver. If 60.0g silver is actually recovered, determine the percent yield of the reactions.

Theoretical yield of silver = 67.97 g Ag

Percent yield of the reaction = 88.3%

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 %

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d: the voltage of the secondary coil is less than that of the primary coil.

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