onsider 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 59.3 g of H2SO4.
Interesting but I don't see a question.
To solve this problem, we need to determine the limiting reactant and calculate the theoretical yield of the reaction.
1. Calculate the molar masses of the reactants and product:
- Molar mass of SO3 (Sulfur trioxide) = 32.07 g/mol (from periodic table)
- Molar mass of H2O (Water) = 18.02 g/mol (from periodic table)
- Molar mass of H2SO4 (Sulfuric acid) = 98.09 g/mol (from periodic table)
2. Determine the moles of each reactant:
- Moles of SO3 = mass / molar mass = 61.5 g / 32.07 g/mol = 1.92 mol
- Moles of H2O = mass / molar mass = 11.2 g / 18.02 g/mol = 0.62 mol
3. Use the balanced equation to determine the stoichiometric ratio between SO3 and H2SO4:
- Balanced equation: SO3(g) + H2O(l) → H2SO4(aq)
- From the equation, we can see that 1 mole of SO3 produces 1 mole of H2SO4.
4. Determine the limiting reactant:
- Since the stoichiometric ratio between SO3 and H2SO4 is 1:1, the reactant that produces the least amount of product is the limiting reactant.
- Comparing the number of moles, we see that H2O has fewer moles (0.62 mol) compared to SO3 (1.92 mol).
- Therefore, H2O is the limiting reactant.
5. Calculate the theoretical yield of H2SO4:
- Theoretical yield = moles of limiting reactant x molar mass of H2SO4
- Theoretical yield = 0.62 mol x 98.09 g/mol = 60.80 g
6. Compare the theoretical yield to the actual yield of H2SO4:
- Actual yield = 59.3 g (given in the problem)
- The actual yield is slightly less than the theoretical yield, which is expected in real-world reactions due to factors such as incomplete reactions and other side reactions.
Overall, the chemist collected 59.3 g of H2SO4, which is slightly less than the calculated theoretical yield of 60.80 g.