Consider the following reaction between sulfur trioxide and water:
SO3(g)+H2O(l)→H2SO4(aq) A chemist allows 61.5 g of SO3 and 11.2 g of H2O to react. When the reaction is finished, the chemist collects 51.0 g of H2SO4. Determine the percent yield for the reaction.
convert mass to moles and see which reagent limits the reaction.
Then compare the 51.0g yield to the theoretical yield using all available materials.
To determine the percent yield for the reaction, we need to compare the actual yield (the amount of H2SO4 collected) to the theoretical yield (the maximum amount of H2SO4 that could be produced according to stoichiometry).
First, let's calculate the molar masses of SO3, H2O, and H2SO4.
Molar mass of SO3:
S: 32.07 g/mol
O: 16.00 g/mol x 3 = 48.00 g/mol
Total: 32.07 g/mol + 48.00 g/mol = 80.07 g/mol
Molar mass of H2O:
H: 1.01 g/mol x 2 = 2.02 g/mol
O: 16.00 g/mol
Total: 2.02 g/mol + 16.00 g/mol = 18.02 g/mol
Molar mass of H2SO4:
H: 1.01 g/mol x 2 = 2.02 g/mol
S: 32.07 g/mol
O: 16.00 g/mol x 4 = 64.00 g/mol
Total: 2.02 g/mol + 32.07 g/mol + 64.00 g/mol = 98.09 g/mol
Next, let's calculate the number of moles of SO3 and H2O used in the reaction.
Number of moles of SO3:
Mass of SO3 = 61.5 g
Molar mass of SO3 = 80.07 g/mol
Moles of SO3 = 61.5 g / 80.07 g/mol = 0.7686 mol (approximately)
Number of moles of H2O:
Mass of H2O = 11.2 g
Molar mass of H2O = 18.02 g/mol
Moles of H2O = 11.2 g / 18.02 g/mol = 0.6214 mol (approximately)
The balanced equation tells us that the stoichiometric ratio between SO3 and H2SO4 is 1:1. This means that 1 mol of SO3 reacts to produce 1 mol of H2SO4.
Since the stoichiometry is 1:1, the number of moles of H2SO4 produced will be equal to the number of moles of SO3 used in the reaction.
Number of moles of H2SO4 = 0.7686 mol
Now, let's calculate the theoretical yield of H2SO4.
Theoretical yield of H2SO4:
Moles of H2SO4 = 0.7686 mol
Molar mass of H2SO4 = 98.09 g/mol
Theoretical yield of H2SO4 = 0.7686 mol x 98.09 g/mol = 75.37 g (approximately)
Finally, let's calculate the percent yield.
Percent yield = (Actual yield / Theoretical yield) x 100
Percent yield = (51.0 g / 75.37 g) x 100 = 67.6% (approximately)
Therefore, the percent yield for the reaction is approximately 67.6%.
To determine the percent yield for the reaction, you need to compare the actual yield (the amount of H2SO4 collected) to the theoretical yield (the maximum amount of H2SO4 that could be produced based on the given amounts of reactants). Follow these steps:
Step 1: Calculate the molar masses of the reactants and the product.
- Molar mass of SO3 (Sulfur Trioxide) = 32.06 g/mol + 3*(16.00 g/mol) = 80.06 g/mol
- Molar mass of H2O (Water) = 2*(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
- Molar mass of H2SO4 (Sulfuric Acid) = 2*(1.01 g/mol) + 32.07 g/mol + 4*(16.00 g/mol) = 98.09 g/mol
Step 2: Convert the given masses of SO3 and H2O to moles.
- Moles of SO3 = 61.5 g / 80.06 g/mol = 0.768 mol
- Moles of H2O = 11.2 g / 18.02 g/mol = 0.621 mol
Step 3: Determine the limiting reagent.
- To find the limiting reagent, compare the mole ratio between SO3 and H2O in the balanced chemical equation (1:1 ratio). The reactant that produces fewer moles of the product is the limiting reagent.
- In this case, both reactants are in excess since their mole ratios are equal.
Step 4: Calculate the theoretical yield of H2SO4.
- Theoretical yield is based on the limiting reagent.
- Since both reactants are in excess, the yield is limited by the amount of SO3.
- From the balanced equation, the molar ratio between SO3 and H2SO4 is 1:1.
- Theoretical yield of H2SO4 = Moles of SO3 = 0.768 mol
Step 5: Calculate the percent yield.
- Percent yield = (Actual yield / Theoretical yield) * 100%
- Actual yield = 51.0 g
- Theoretical yield = 0.768 mol * 98.09 g/mol = 75.4 g
- Percent yield = (51.0 g / 75.4 g) * 100% = 67.6%
Therefore, the percent yield for the reaction is 67.6%.