How many millilitres of ozone gas at 25.0 °C and 1.00 bar are needed to react with 45.00 mL of a 0.100 mol L^-1 aqueous solution of KI according to the chemical equation shown below?
O₃(g) + 2I^-(aq) + H₂O(l) → O₂(g) + I₂(s) + 2OH^-(aq)

Question

Quizplus · Accepted Answer

The reaction requires 1 mole of O3 for every 2 moles of I-. Calculate moles of I-: 45.00 mL x 0.100 mol/L = 0.0045 mol. Moles of O3 needed: 0.0045 mol / 2 = 0.00225 mol. Using ideal gas law at 25°C and 1 bar, volume of O3 = 0.00225 mol x 24.8 L/mol = 0.0558 L = 55.8 mL, closest to 54.8 mL.

How Many Millilitres of Ozone Gas at 25