I don't know what a copper cycle lab is.
a. mols Cu(NO3)2 = M x L = ?
1 mol Cu in 1 mol Cu(NO3)2
g Cu = mols x atomic mass = theoretical yield
b. 0.075 from the problem.
c. (actual/theoretical)*100 = %yield
A) What is the Theoretical Yield
B) What is the Actual Yield
C) What is the Percent Yield
a. mols Cu(NO3)2 = M x L = ?
1 mol Cu in 1 mol Cu(NO3)2
g Cu = mols x atomic mass = theoretical yield
b. 0.075 from the problem.
c. (actual/theoretical)*100 = %yield
The balanced chemical equation for the reaction is as follows:
2Cu(NO3)2 + 2NaOH → Cu(OH)2 + 2NaNO3
A) The Theoretical Yield represents the maximum amount of product that can be obtained from the given reactants. To find the theoretical yield, we use stoichiometry.
1 mole of Cu(NO3)2 reacts to produce 1 mole of Cu(OH)2. The molar mass of Cu(NO3)2 is (63.55 g/mol + 14.01 g/mol + (3 × 16.00 g/mol)) = 187.56 g/mol.
To find the number of moles of Cu(NO3)2 in 9.7 mL of a 0.26 M solution, we use the formula:
moles = volume (mL) × concentration (M) / 1000
moles = 9.7 mL × 0.26 M / 1000 = 0.002522 moles
Since the stoichiometric ratio is 1:1, the theoretical yield of Cu(OH)2 is also 0.002522 moles.
To convert moles to grams, we multiply by the molar mass of Cu(OH)2. The molar mass of Cu(OH)2 is (63.55 g/mol + 16.00 g/mol + (2 × 1.01 g/mol)) = 97.56 g/mol.
The theoretical yield of Cu(OH)2 is therefore:
theoretical yield = 0.002522 moles × 97.56 g/mol = 0.2459 grams.
B) The Actual Yield is the amount of product actually obtained in the experiment. In this case, the student recovers 0.075 grams of copper metal.
C) The Percent Yield is the ratio of the actual yield to the theoretical yield, multiplied by 100 to get the percentage:
percent yield = (actual yield / theoretical yield) × 100
percent yield = (0.075 g / 0.2459 g) × 100 = 30.52%
So, the answers to the questions are:
A) The Theoretical Yield is 0.2459 grams
B) The Actual Yield is 0.075 grams
C) The Percent Yield is 30.52%