When copper wire is placed into a silver nitrate solution, silver crystals and copper(II) nitrate solution form. If 57.1 g silver is actually recovered from the reaction, determine the percent yield of the reaction.

Question

When copper wire is placed into a silver nitrate solution, silver crystals and copper(II) nitrate solution form. If 57.1 g silver is actually recovered from the reaction, determine the percent yield of the reaction.

Answer 1

2AgNO3 + Cu ==>Cu(NO3)2 + 2Ag

Here is an example of a stoichiometry problem (solved). Just follow the steps. You can calculate the theoretical yield in that manner, then %yield = 57.1 g Ag/theoretical yield)*100 = ??
http://www.jiskha.com/science/chemistry/stoichiometry.html

Answer 2

To determine the percent yield of the reaction, you need to compare the actual yield (57.1 g of silver) to the theoretical yield. The theoretical yield is the maximum amount of product that can be formed based on stoichiometry and assuming no losses during the reaction.

First, you need to balance the chemical equation for the reaction:

Cu + 2AgNO3 → 2Ag + Cu(NO3)2

From the balanced equation, you can see that the mole ratio between Cu and Ag is 1:2. So, according to stoichiometry, for every 1 mole of Cu used, 2 moles of Ag are produced.

To calculate the theoretical yield of silver, you need to convert the mass of Cu used to moles and then use the mole ratio to find the moles of Ag produced. Finally, convert the moles of Ag to grams.

Now, let's calculate the theoretical yield:

1. Convert the mass of Cu (given in the problem) to moles by dividing by the molar mass of Cu (63.55 g/mol):

57.1 g Cu × (1 mol Cu / 63.55 g Cu) = 0.897 mol Cu

2. Use the mole ratio from the balanced equation to find the moles of Ag produced:

0.897 mol Cu × (2 mol Ag / 1 mol Cu) = 1.794 mol Ag

3. Convert the moles of Ag to grams by multiplying by the molar mass of Ag (107.87 g/mol):

1.794 mol Ag × (107.87 g Ag / 1 mol Ag) = 193.57 g Ag (theoretical yield)

Now that we have the theoretical yield (193.57 g Ag) and the actual yield (57.1 g Ag), we can calculate the percent yield:

Percent yield = (actual yield / theoretical yield) × 100%
Percent yield = (57.1 g Ag / 193.57 g Ag) × 100%
Percent yield = 29.5%

Therefore, the percent yield of the reaction is approximately 29.5%.