One of the reactions involved in the smelting of copper sulfide ores involves copper(I) oxide and copper (I) sulfide: 2Cu2O(solid) + Cu2S(solid) ---> 6Cu(solid) + SO2(gas) 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
mols Cu2O = grams/molar mass = ?
mols Cu2S = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Cu2O to mols Cu.
Do the same for mols Cu2S to mols Cu.
It is likely these two values will not be the same; in limiting reagent problems the smaller number is ALWAYS the right value to choose and the reagent producing that value is the limiting reagent. The other reagent will be the one in excess.
Using the smaller value, convert mols to grams of the product. g = mols x atomic mass Cu. This is the theoretical yield (TY). The actual yield (AY) is 35.46 from the problem.
% yield = (AY/TY)*100 = ?
To determine which reagent is present in excess, we need to calculate the number of moles of each reagent. We can do this by using the molar mass and the given mass of each reagent. Here are the steps to follow:
a) Determine which reagent is present in excess:
Step 1: Calculate the number of moles of copper(I) oxide (Cu2O):
Number of moles = Mass / Molar mass
Molar mass of Cu2O = 2(Atomic mass of Cu) + Atomic mass of O
(Atomic mass of Cu is 63.55 g/mol, and Atomic mass of O is 16.00 g/mol)
Number of moles of Cu2O = 35.00 g / (2 * 63.55 g/mol + 16.00 g/mol)
Step 2: Calculate the number of moles of copper(I) sulfide (Cu2S):
Number of moles = Mass / Molar mass
Molar mass of Cu2S = 2(Atomic mass of Cu) + Atomic mass of S
(Atomic mass of S is 32.07 g/mol)
Number of moles of Cu2S = 25.00 g / (2 * 63.55 g/mol + 32.07 g/mol)
Step 3: Compare the number of moles of each reagent. The one with less number of moles is the limiting reagent, and the one with more number of moles is in excess.
Let's assume the number of moles of Cu2O is "x", and the number of moles of Cu2S is "y".
If "x" is less than "y", then Cu2O is the limiting reagent.
If "x" is greater than "y", then Cu2S is the limiting reagent.
b) Calculate the theoretical yield of copper:
Step 1: Determine the stoichiometry of the balanced equation.
From the balanced equation: 2Cu2O + Cu2S --> 6Cu + SO2
We can see that the ratio of Cu2O to Cu is 2:6, or simplified to 1:3.
Step 2: Calculate the moles of Cu produced from the limiting reagent.
Multiply the moles of the limiting reagent (either Cu2O or Cu2S) by the stoichiometric ratio.
Theoretical moles of Cu = Moles of limiting reagent * (6 moles of Cu / 2 moles of Cu2O or Cu2S)
Theoretical moles of Cu = (moles of Cu2O or Cu2S) * 3
Step 3: Calculate the theoretical yield of Cu in grams.
Theoretical yield of Cu = Theoretical moles of Cu * Molar mass of Cu
c) Determine the percentage yield:
Step 1: Calculate the actual yield of copper in grams, given as 35.46 g.
Step 2: Calculate the percentage yield using the formula:
Percentage yield = (Actual yield / Theoretical yield) * 100%
Please provide the actual yield of copper so we can complete step 1 and calculate the percentage yield in step 2.
To determine which reagent is present in excess, we can compare the stoichiometric ratios of the two reactants.
a) Calculate the number of moles for each reactant:
- Copper(I) oxide (Cu2O):
- Mass = 35.00 g
- Molar mass of Cu2O = 2 * atomic mass of Cu + atomic mass of O = 2 * (63.55 g/mol) + 16.00 g/mol = 143.10 g/mol
- Moles = mass / molar mass = 35.00 g / 143.10 g/mol = 0.2442 mol
- Copper(I) sulfide (Cu2S):
- Mass = 25.00 g
- Molar mass of Cu2S = 2 * atomic mass of Cu + atomic mass of S = 2 * (63.55 g/mol) + 32.07 g/mol = 159.17 g/mol
- Moles = mass / molar mass = 25.00 g / 159.17 g/mol = 0.1571 mol
Since there is a 2:1 stoichiometric ratio between Cu2O and Cu2S, we can see that Copper(I) oxide is in excess because we have a greater number of moles (0.2442 mol) of Copper(I) oxide compared to Copper(I) sulfide (0.1571 mol).
b) To calculate the theoretical yield of copper, we need to determine which reactant limits the reaction. Since Copper(I) oxide is in excess, we can use the moles of Copper(I) sulfide to calculate the theoretical yield of copper.
- From the balanced equation, we can see that the stoichiometric ratio is 2:6 for Copper(I) sulfide to Copper.
- The number of moles of copper formed would be 6 moles for every 0.1571 mol of Copper(I) sulfide.
- Therefore, moles of copper = 6 * 0.1571 mol = 0.9426 mol
Now, we can calculate the mass of copper using the molar mass of copper:
- Molar mass of Cu = 63.55 g/mol
- Mass of copper = moles of copper * molar mass of Cu = 0.9426 mol * 63.55 g/mol = 59.92 g
Therefore, the theoretical yield of copper is 59.92 g.
c) To calculate the percentage yield, we compare the actual yield (35.46 g) with the theoretical yield (59.92 g).
- Percentage yield = (actual yield / theoretical yield) * 100
- Percentage yield = (35.46 g / 59.92 g) * 100 = 59.20%
Therefore, the percentage yield of copper is 59.20%.