Calculate the percent yield for an experiment in which 46.9 g of SO3 was produced upon the reaction of 51.2 g of SO2 with 25.0 g O2
2SO2 + O2 ==> 2SO3
mols SO2 = grams/molar mass
mols O2 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols SO2 to mols SO3
Do the same for mols O2 to mols SO3.
It is likely these values will be different which means one of them is wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Now convert the smaller value to g SO3. g = mols x molar mass. This is the theoretical yield (TY). The actual yield (AY) from the problem is 46.9.
Then %yield = (AY/TY)*100 = ?
Well, let's do some calculations... *drumroll*
The first thing we need to do is determine the theoretical yield, which is the maximum amount of product that can be obtained from the given reactants.
To find the theoretical yield, we need to compare the stoichiometric ratios of the reactants and the product. The balanced equation for the reaction is:
2 SO2 + O2 → 2 SO3
From the equation, we can see that 2 moles of SO2 react with 1 mole of O2 to produce 2 moles of SO3. Now we can calculate the amount of SO3 that can be produced from the given amounts of SO2 and O2:
Molar mass of SO2 = 32.00 g/mol
Molar mass of O2 = 32.00 g/mol
Moles of SO2 = 51.2 g / 32.00 g/mol = 1.60 mol
Moles of O2 = 25.0 g / 32.00 g/mol = 0.78 mol
Since the stoichiometric ratio is 2:1, we have excess SO2 left over. So, we can only produce the same amount of SO3 as the limiting reactant, which is O2.
Moles of SO3 = 0.78 mol
Now, let's calculate the molar mass of SO3:
Molar mass of SO3 = 32.06 g/mol (approximately)
Theoretical yield of SO3 = Moles of SO3 × Molar mass of SO3
= 0.78 mol × 32.06 g/mol
= 24.99 g
Now we have the theoretical yield, which is 24.99 g. The actual yield is given as 46.9 g.
Percent yield = (Actual yield / Theoretical yield) × 100
= (46.9 g / 24.99 g) × 100
≈ 187.69%
Now that's an impressive percent yield! It seems like you got more product than even the theoretical yield. It's like finding extra fries in your fast-food bag. Well done!
To calculate the percent yield, we need to compare the actual yield (the amount of product obtained from the reaction) with the theoretical yield (the maximum amount of product that could be obtained, based on stoichiometry and assuming 100% efficiency).
First, we need to determine the limiting reactant of the reaction. This is the reactant that will be completely consumed, and it determines the maximum amount of product that can be formed.
To find the limiting reactant, we will compare the moles of each reactant. The balanced equation for the reaction is:
2 SO2 + O2 -> 2 SO3
First, we convert the given masses of SO2, O2, and SO3 to moles using their molar masses.
Molar mass of SO2: 32.06 g/mol
Molar mass of O2: 31.999 g/mol
Molar mass of SO3: 80.06 g/mol
Moles of SO2 = 51.2 g / 32.06 g/mol = 1.597 mol
Moles of O2 = 25.0 g / 31.999 g/mol = 0.781 mol
Next, we compare the moles of each reactant to the stoichiometric ratio in the balanced equation. In this case, the ratio is 2:1 for SO2 to O2, so we need to convert the moles of O2 to the equivalent moles of SO2:
Moles of O2 required = 2 * Moles of SO2 = 2 * 1.597 mol = 3.194 mol
Since 0.781 mol of O2 is less than the required 3.194 mol, O2 is the limiting reactant.
Now we can calculate the theoretical yield of SO3 using the moles of the limiting reactant.
Moles of SO3 = 0.781 mol of O2 * (2 mol of SO3 / 1 mol of O2) = 1.562 mol
Finally, we can calculate the theoretical yield of SO3 in grams using the molar mass of SO3.
Theoretical yield of SO3 = 1.562 mol * 80.06 g/mol = 125.01 g
Now we can calculate the percent yield using the actual and theoretical yields.
Percent yield = (actual yield / theoretical yield) * 100
Actual yield = 46.9 g
Percent yield = (46.9 g / 125.01 g) * 100 = 37.5%
Therefore, the percent yield for this experiment is 37.5%.
To calculate the percent yield, we need to compare the actual yield (the amount of SO3 produced) to the theoretical yield (the amount of SO3 that should have been produced if the reaction had gone to completion).
First, we need to determine the balanced chemical equation for the reaction:
2 SO2 + O2 -> 2 SO3
Next, we need to calculate the theoretical yield of SO3 based on the given mass of SO2:
Step 1: Convert the mass of SO2 to moles using the molar mass of SO2.
Molar mass of SO2 = 32.07 g/mol + 32.07 g/mol = 64.14 g/mol
Moles of SO2 = mass of SO2 / molar mass of SO2
= 51.2 g / 64.14 g/mol
= 0.7989 mol SO2
Step 2: Use stoichiometry to determine the moles of SO3 that should have been produced. From the balanced equation, we know that the ratio of moles of SO2 to moles of SO3 is 2:2 (or 1:1).
Moles of SO3 = moles of SO2 (from step 1) = 0.7989 mol SO3
Step 3: Convert the moles of SO3 to mass using the molar mass of SO3.
Molar mass of SO3 = 32.07 g/mol + 32.07 g/mol + 16.00 g/mol = 80.14 g/mol
Mass of SO3 = moles of SO3 * molar mass of SO3
= 0.7989 mol * 80.14 g/mol
= 63.99 g
The theoretical yield of SO3 should be 63.99 g.
Now, we can calculate the percent yield using the formula:
Percent yield = (Actual yield / Theoretical yield) * 100
Given that the actual yield of SO3 is 46.9 g, we can plug in the values:
Percent yield = (46.9 g / 63.99 g) * 100
= 73.46%
Therefore, the percent yield for this experiment is approximately 73.46%.