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 a determine the limiting reactant for the reaction. express your answer as a chemical formula.

Explanation:

Step1: Calculate moles of \( SO_3 \)

Molar mass of \( SO_3 \) is \( 32.07 + 3\times16.00 = 80.07 \, g/mol \).
Moles of \( SO_3 = \frac{61.5 \, g}{80.07 \, g/mol} \approx 0.768 \, mol \).

Step2: Calculate moles of \( H_2O \)

Molar mass of \( H_2O \) is \( 2\times1.008 + 16.00 = 18.016 \, g/mol \).
Moles of \( H_2O = \frac{11.2 \, g}{18.016 \, g/mol} \approx 0.622 \, mol \).

Step3: Determine limiting reactant

From the reaction \( SO_3(aq) + H_2O(l)
ightarrow H_2SO_4(aq) \), the mole ratio of \( SO_3 \) to \( H_2O \) is \( 1:1 \).
Since moles of \( H_2O (0.622 \, mol) < \) moles of \( SO_3 (0.768 \, mol) \), \( H_2O \) is the limiting reactant.

Answer:

\( H_2O \)