Sulfur trioxide, SO3, is made from the oxidation of SO2, and the reaction is represented by the equation
2SO2 + O2 2SO3
A 25-g sample of SO2 gives 18 g of SO3. The PERCENT YIELD of SO3 is?
To find the percent yield of SO3, we need to compare the actual yield (the amount we obtained in the experiment) with the theoretical yield (the amount we expect to obtain based on the given equation).
First, we need to determine the theoretical yield of SO3. We can use stoichiometry to calculate this value.
From the balanced equation:
2SO2 + O2 → 2SO3
We can see that 2 moles of SO2 produce 2 moles of SO3. Therefore, the molar ratio between SO3 and SO2 is 1:1.
Given that the molar mass of SO2 is approximately 64 g/mol, we can calculate the number of moles of SO2 in the 25 g sample using the formula:
moles of SO2 = mass of SO2 / molar mass of SO2
moles of SO2 = 25 g / 64 g/mol
moles of SO2 ≈ 0.391 mol
Since the molar ratio between SO3 and SO2 is 1:1, the theoretical yield of SO3 would also be approximately 0.391 mol.
To convert the moles of SO3 to grams, we use the molar mass of SO3. The molar mass of SO3 is approximately 80 g/mol.
The theoretical yield of SO3 in grams is:
theoretical yield of SO3 = moles of SO3 × molar mass of SO3
theoretical yield of SO3 = 0.391 mol × 80 g/mol
theoretical yield of SO3 ≈ 31.3 g
Now we can calculate the percent yield of SO3 using the formula:
percent yield = (actual yield / theoretical yield) × 100
Given that the actual yield of SO3 is 18 g:
percent yield = (18 g / 31.3 g) × 100
percent yield ≈ 57.5%
Therefore, the percent yield of SO3 is approximately 57.5%.
To calculate the percent yield of SO3, we need to compare the actual yield (the amount we obtained experimentally) with the theoretical yield (the amount we would expect to obtain if the reaction went to completion).
First, let's determine the theoretical yield of SO3 based on the balanced equation.
From the equation: 2SO2 + O2 -> 2SO3
The molar ratio tells us that 2 moles of SO2 react with 1 mole of O2 to produce 2 moles of SO3.
Next, we need to find the moles of SO2 in the 25 g sample. We can use the molar mass of SO2 to convert grams to moles.
Molar mass of SO2 = 2 * (Sulfur's atomic mass) + 2 * (Oxygen's atomic mass)
= 2 * (32.07 g/mol) + 2 * (16.00 g/mol)
= 64.14 g/mol + 32.00 g/mol
= 96.14 g/mol
Moles of SO2 = Mass of SO2 / Molar mass of SO2
= 25 g / 96.14 g/mol
≈ 0.26 mol
According to the stoichiometry, 2 moles of SO2 react to produce 2 moles of SO3.
Therefore, if 0.26 moles of SO2 react completely, the theoretical yield of SO3 would be 0.26 mol.
Now, we can calculate the percent yield.
Percent yield = (Actual yield / Theoretical yield) * 100
Given that the actual yield of SO3 is 18 g, we need to convert grams to moles using the molar mass.
Molar mass of SO3 = 1 * (Sulfur's atomic mass) + 3 * (Oxygen's atomic mass)
= 1 * (32.07 g/mol) + 3 * (16.00 g/mol)
= 32.07 g/mol + 48.00 g/mol
= 80.07 g/mol
Moles of SO3 = Mass of SO3 / Molar mass of SO3
= 18 g / 80.07 g/mol
≈ 0.2248 mol
Now we can calculate the percent yield:
Percent yield = (0.2248 mol / 0.26 mol) * 100
≈ 86.5%
Therefore, the percent yield of SO3 is approximately 86.5%.
You know from your first post about 7.0 g propane how to do stoichiometry. The number you obtained in that process is called the theoretical yield (if you had determined either CO2 or H2O). Do this one the same way. The answer you obtain will be the theoretical yield. Then
%yield = (actual yield/theoretical yield)*100 = ??
The actual yield is given in the problem as 18 g SO3.