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 59.0 g of H2SO4.
I don't see a question here. However, this is a limiting reagent problem.
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To find the limiting reactant in a chemical reaction, we need to compare the number of moles of each reactant with the stoichiometric coefficients in the balanced equation. The reactant that produces the smallest number of moles of the product is the limiting reactant, as it will be completely consumed first.
First, we need to calculate the number of moles of each reactant using their respective molar masses. The molar mass of SO3 (sulfur trioxide) is calculated by adding the atomic masses of sulfur (S) and three times the atomic mass of oxygen (O):
Molar mass of SO3 = atomic mass of S + 3 × atomic mass of O
= 32.07 g/mol + 3 × 16.00 g/mol
= 80.07 g/mol
Next, we can calculate the number of moles of SO3:
Number of moles of SO3 = mass of SO3 / molar mass of SO3
= 61.5 g / 80.07 g/mol
≈ 0.768 moles
Similarly, we calculate the number of moles of H2O:
Molar mass of H2O = atomic mass of H + 2 × atomic mass of O
= 1.01 g/mol + 2 × 16.00 g/mol
= 18.02 g/mol
Number of moles of H2O = mass of H2O / molar mass of H2O
= 11.2 g / 18.02 g/mol
≈ 0.621 moles
Now, we can use the balanced equation to determine the stoichiometric ratio between SO3 and H2O. From the equation:
1 mole of SO3 reacts with 1 mole of H2O to produce 1 mole of H2SO4.
Comparing the molar ratios, we find that the number of moles of SO3 is equal to the number of moles of H2O.
Since the reaction requires an equal number of moles of SO3 and H2O, we can conclude that both SO3 and H2O are in a 1:1 stoichiometric ratio in this reaction.
However, we are also given the mass of H2SO4 produced. To find the percent yield of the reaction, we compare the actual yield (mass of H2SO4 collected) with the theoretical yield (calculated using the limiting reactant):
Theoretical yield of H2SO4 = (molar mass of H2SO4) x (number of moles of limiting reactant)
Molar mass of H2SO4 = 1.01 g/mol (atomic mass of H) + 2 × 16.00 g/mol (atomic mass of O) + 32.07 g/mol (atomic mass of S)
= 98.09 g/mol
Number of moles of limiting reactant (both SO3 and H2O have the same number of moles) = 0.621 moles
Theoretical yield of H2SO4 = 98.09 g/mol × 0.621 moles
≈ 60.88 g
Finally, we can calculate the percent yield:
Percent yield = (actual yield / theoretical yield) × 100
= (59.0 g / 60.88 g) × 100
≈ 96.7%
Therefore, the chemist obtained approximately a 96.7% yield of H2SO4 from the reaction.