For the reaction SO3 plus H2O>>H2SO4, calculate the percent yield if 500.g of sulfur trioxide react with excess water to produce 575g of sulfuric acid??
for each mole of sulfur trioxide, one gets one mole of sulfuric acid.
93.9%
To calculate the percent yield, we need to compare the actual yield (the amount of product obtained in the reaction) with the theoretical yield (the maximum amount of product that can be obtained according to stoichiometry).
1. Calculate the molar mass of SO3:
S = 32.07 g/mol
O = 16.00 g/mol
Molar mass of SO3 = (32.07 g/mol) + 3*(16.00 g/mol) = 80.07 g/mol
2. Calculate the number of moles of SO3:
Moles of SO3 = mass / molar mass
Moles of SO3 = 500 g / 80.07 g/mol = 6.247 mol
3. In the balanced chemical equation, the stoichiometry between SO3 and H2SO4 is 1:1. So, the number of moles of H2SO4 produced will be the same as the number of moles of SO3.
4. Calculate the theoretical yield of H2SO4:
Theoretical yield = Moles of H2SO4 * molar mass of H2SO4
Molar mass of H2SO4 = (2*1.01 g/mol) + 32.07 g/mol + 4*(16.00 g/mol) = 98.09 g/mol
Theoretical yield = 6.247 mol * 98.09 g/mol = 612.19 g
5. Calculate the percent yield:
Percent yield = (Actual yield / Theoretical yield) * 100
Actual yield = 575 g
Percent yield = (575 g / 612.19 g) * 100 = 93.9%
Therefore, the percent yield for the reaction is approximately 93.9%.
To calculate the percent yield for the reaction, we need to compare the actual yield with the theoretical yield. The theoretical yield is the maximum amount of product that can be formed, assuming 100% efficiency in the reaction.
First, let's calculate the molar masses of the reactants and the product:
- The molar mass of SO3 (sulfur trioxide) is 32.06 g/mol (sulfur: 32.06 g/mol, oxygen: 3 x 16.00 g/mol).
- The molar mass of H2O (water) is 18.02 g/mol (hydrogen: 2 x 1.01 g/mol, oxygen: 16.00 g/mol).
- The molar mass of H2SO4 (sulfuric acid) is 98.09 g/mol (hydrogen: 2 x 1.01 g/mol, sulfur: 32.06 g/mol, oxygen: 4 x 16.00 g/mol).
Next, we need to convert the masses of the reactants and product into moles:
- Moles of SO3 = mass of SO3 / molar mass of SO3
= 500 g / 32.06 g/mol
- Moles of H2SO4 = mass of H2SO4 / molar mass of H2SO4
= 575 g / 98.09 g/mol
Now, we can calculate the mole ratio of SO3 to H2SO4 using the balanced equation:
- From the balanced equation, we can see that the mole ratio is 1:1. This means that for every 1 mole of SO3, 1 mole of H2SO4 is produced.
Now, we can calculate the theoretical yield of H2SO4:
- Theoretical yield of H2SO4 = Moles of H2SO4
= Moles of SO3 (since the mole ratio is 1:1)
Finally, we can calculate the percent yield:
- Percent yield = (Actual yield / Theoretical yield) x 100%
= (575 g / Theoretical yield) x 100%