Question

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 53.7 g of h₂so₄. part b determine the theoretical yield for the reaction. express your answer to three significant figures.

Explanation:

Step1: Find moles of reactants

Molar mass of \( \text{SO}_3 \): \( 32.07 + 3\times16.00 = 80.07 \, \text{g/mol} \)
Moles of \( \text{SO}_3 \): \( \frac{61.5 \, \text{g}}{80.07 \, \text{g/mol}} \approx 0.768 \, \text{mol} \)

Molar mass of \( \text{H}_2\text{O} \): \( 2\times1.01 + 16.00 = 18.02 \, \text{g/mol} \)
Moles of \( \text{H}_2\text{O} \): \( \frac{11.2 \, \text{g}}{18.02 \, \text{g/mol}} \approx 0.621 \, \text{mol} \)

Step2: Determine limiting reactant

Reaction: \( \text{SO}_3 + \text{H}_2\text{O}
ightarrow \text{H}_2\text{SO}_4 \) (1:1 mole ratio)
Since \( 0.621 < 0.768 \), \( \text{H}_2\text{O} \) is limiting.

Step3: Calculate theoretical yield (moles of \( \text{H}_2\text{SO}_4 \))

Moles of \( \text{H}_2\text{SO}_4 \) = moles of limiting reactant (\( \text{H}_2\text{O} \)) = \( 0.621 \, \text{mol} \)

Step4: Convert moles to grams

Molar mass of \( \text{H}_2\text{SO}_4 \): \( 2\times1.01 + 32.07 + 4\times16.00 = 98.09 \, \text{g/mol} \)
Theoretical yield: \( 0.621 \, \text{mol} \times 98.09 \, \text{g/mol} \approx 60.9 \, \text{g} \)

Answer:

60.9