Copper(II) sulfide reacts with oxygen to produce copper (II) oxide plus sulfur dioxide.
Suppose you start this reaction with 50.0 grams of copper(II) sulfide and 50.0 grams of oxygen and you actually produce 33.5 grams of copper(II) oxide.
Calculate the percent yield for the reaction.
This is a limiting reagent (LR) problem. You know that because amounts are given for BOTH reactants.
2CuS + 3O2 ==> 2CuO + 2SO2
mols CuS = grams/molar mass
mols O2 = grams/molar mass
Using the coefficients in the balanced equation, convert mols CuS to mols CuO.
Do the same for mols O2 to mols CuO.
It is likely these to values for CuO will not agree which means one is not right; the correct answer in LR problem is ALWAYS the smaller value and the reagent producing that value is the LR.
Using the smaller value, convert mols CuO to grams. g = mols x molar mass. This is the theoretical yield(TY). The actual yield (AY) is given in the problem as 33.5 g.
% yield = (AY/TY)*100 = ?
To calculate the percent yield for the reaction, we first need to determine the theoretical yield and the actual yield.
1. Calculate the molar masses:
- Copper(II) sulfide (CuS): 63.55 g/mol (atomic mass of Cu) + 32.07 g/mol (atomic mass of S) = 95.62 g/mol
- Oxygen (O₂): 16.00 g/mol (atomic mass of O) × 2 = 32.00 g/mol
- Copper(II) oxide (CuO): 63.55 g/mol (atomic mass of Cu) + 16.00 g/mol (atomic mass of O) = 79.55 g/mol
2. Calculate the moles of each reactant:
- Moles of CuS = mass / molar mass = 50.0 g / 95.62 g/mol = 0.523 mol
- Moles of O₂ = mass / molar mass = 50.0 g / 32.00 g/mol = 1.563 mol
3. Use the balanced chemical equation to determine the stoichiometric ratio:
CuS + O₂ → CuO + SO₂
From the balanced equation, we can see that 1 mole of CuS reacts with 1 mole of O₂ to produce 1 mole of CuO.
4. Determine the limiting reactant:
The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed. To find the limiting reactant, compare the moles of the reactants to their stoichiometric ratios in the balanced equation.
- For CuS: 0.523 mol of CuS × (1 mol CuO / 1 mol CuS) = 0.523 mol of CuO
- For O₂: 1.563 mol of O₂ × (1 mol CuO / 1 mol O₂) = 1.563 mol of CuO
Since we have less moles of CuS (0.523 mol) compared to O₂ (1.563 mol), CuS is the limiting reactant.
5. Calculate the theoretical yield of CuO, using the stoichiometry:
Theoretical yield of CuO = moles of limiting reactant × molar mass of CuO
Theoretical yield of CuO = 0.523 mol × 79.55 g/mol = 41.58 g
6. Calculate the percent yield:
Percent yield = (actual yield / theoretical yield) × 100%
Percent yield = (33.5 g / 41.58 g) × 100% = 80.63%
Therefore, the percent yield for the reaction is 80.63%.
To calculate the percent yield, you need to compare the actual yield (33.5 grams of copper(II) oxide) to the theoretical yield (the maximum amount of copper(II) oxide that could be produced from the given reactants).
First, we need to determine the balanced chemical equation for the reaction:
2 CuS + 3 O2 -> 2 CuO + 2 SO2
From the balanced equation, we can see that 2 moles of CuS react with 3 moles of O2 to produce 2 moles of CuO.
Next, let's calculate the molar masses of the compounds involved:
- CuS (copper(II) sulfide) has a molar mass of 95.61 g/mol. (Atomic mass of copper = 63.55 g/mol, atomic mass of sulfur = 32.07 g/mol)
- O2 (oxygen) has a molar mass of 32.00 g/mol.
- CuO (copper(II) oxide) has a molar mass of 79.55 g/mol.
Now we can determine the theoretical yield of CuO:
1. Calculate the moles of CuS:
Moles of CuS = mass / molar mass = 50.0 g / 95.61 g/mol = 0.5228 mol
2. Calculate the moles of O2:
Moles of O2 = mass / molar mass = 50.0 g / 32.00 g/mol = 1.5625 mol
3. Determine the limiting reactant:
The limiting reactant is the one that is completely consumed and determines the maximum amount of product formed. In this case, we need to compare the moles of CuS and O2:
- Since the ratio of CuS to O2 in the balanced equation is 2:3, we can use the mole ratio to determine the number of moles of CuS required to react with the given moles of O2.
- If we multiply the moles of O2 by 2/3 (moles of CuS needed per mole of O2), we can compare that value to the actual moles of CuS to determine the limiting reactant.
Moles of CuS required = (1.5625 mol O2) * (2 mol CuS / 3 mol O2) = 1.0417 mol
Since we have only 0.5228 mol of CuS, it is the limiting reactant.
4. Calculate the theoretical yield of CuO using the moles of CuS:
Theoretical yield of CuO = (moles of CuS) * (molar mass of CuO) = 0.5228 mol * 79.55 g/mol = 41.57 g
Finally, we can calculate the percent yield using the actual yield and theoretical yield:
Percent yield = (actual yield / theoretical yield) * 100
Percent yield = (33.5 g / 41.57 g) * 100 = 80.67%
Therefore, the percent yield for the reaction is approximately 80.67%.