One of the reactions involved in the smelting of copper sulfide ores involves copper (I) oxide and copper (I) sulfide: 2Cu2O2 + Cu2S ---> 6Cu + SO2
Assuming that 35.00 g of copper (I) oxide is heated with 25.00 g of copper (I) sulfide
A) determine which reagent is present in excess
B) calculate the theoretical yield of copper
C) determine the percentage yield if 35.46 g of copper is actually isolated
My work
A) Cu2O = 0.2446
Cu2S = 0.1571
Not sure how to do B) and C)
http://www.jiskha.com/display.cgi?id=1417920941
To determine which reagent is present in excess, you need to compare the amount of each reagent present in the reaction.
A) We can find the moles of each reagent using its molar mass:
For copper(I) oxide (Cu2O):
Molar mass = 2*(63.55 g/mol) + 16.00 g/mol = 143.10 g/mol
Moles = Mass / Molar mass = 35.00 g / 143.10 g/mol
For copper(I) sulfide (Cu2S):
Molar mass = 2*(63.55 g/mol) + 32.07 g/mol = 159.17 g/mol
Moles = Mass / Molar mass = 25.00 g / 159.17 g/mol
Now, divide the moles of each reagent by their stoichiometric coefficients from the balanced equation:
Moles of Cu2O / Coefficient = 0.2446 mol / 2 = 0.1223 mol
Moles of Cu2S / Coefficient = 0.1571 mol / 1 = 0.1571 mol
Comparing the moles, we see that copper(II) sulfide (Cu2S) is present in excess.
B) To calculate the theoretical yield of copper (Cu), we need to determine the limiting reagent and use its stoichiometric coefficient from the balanced equation.
From the previous calculations, we established that Cu2S is the excess reagent. Therefore, Cu2O is the limiting reagent.
The balanced equation provides a 2:6 molar ratio between Cu2O and Cu. This means that for every 2 moles of Cu2O, we should obtain 6 moles of Cu.
Moles of Cu = Moles of Cu2O * (6/2) = 0.1223 mol * (6/2) = 0.3669 mol
To convert moles of Cu to grams, we use its molar mass:
Molar mass of Cu = 63.55 g/mol
Mass of Cu = Moles of Cu * Molar mass of Cu = 0.3669 mol * 63.55 g/mol
Mass of Cu = 23.32 g
Therefore, the theoretical yield of copper is 23.32 grams.
C) To determine the percentage yield, we need to compare the actual yield of copper to the theoretical yield we calculated.
Actual yield of copper = 35.46 g (given in the question)
Percentage yield = (Actual yield / Theoretical yield) * 100
Percentage yield = (35.46 g / 23.32 g) * 100
Percentage yield ≈ 152%
The percentage yield is approximately 152%. This value suggests that the actual yield exceeds the theoretical yield, which can sometimes occur due to experimental errors or incomplete reactions.