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. determine the percent yield for the reaction. express your answer to three significant figures.

Explanation:

Step1: Find moles of reactants

Molar mass of \(SO_3\): \(32.07 + 3\times16.00 = 80.07\) g/mol. Moles of \(SO_3\): \(\frac{51.5\ g}{80.07\ g/mol} \approx 0.643\) mol.
Molar mass of \(H_2O\): \(2\times1.01 + 16.00 = 18.02\) g/mol. Moles of \(H_2O\): \(\frac{11.2\ g}{18.02\ g/mol} \approx 0.621\) mol.

Step2: Determine limiting reactant

Reaction: \(SO_3(g) + H_2O(l)
ightarrow H_2SO_4(aq)\). Molar ratio \(1:1\). \(H_2O\) has fewer moles (0.621 < 0.643), so \(H_2O\) is limiting.

Step3: Calculate theoretical yield of \(H_2SO_4\)

Molar mass of \(H_2SO_4\): \(2\times1.01 + 32.07 + 4\times16.00 = 98.09\) g/mol.
Theoretical yield: \(0.621\ mol \times 98.09\ g/mol \approx 60.9\) g.

Step4: Calculate percent yield

Percent yield = \(\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 = \frac{53.7\ g}{60.9\ g} \times 100 \approx 88.2\%\).

Answer:

\(88.2\%\)