How many milliliters of ozone gas at at 25.0°C and 1.00 atm pressure are needed to react with 45.00 mL of a 0.200 M aqueous solution of KI according to the chemical equation shown below? O₃(g)+ 2 I^-(aq)+ H₂O(l)→ O₂(g)+ I₂(s)+ 2 OH^-(aq) A) 110. mL B) 220. mL C) 330. mL D) 441 mL

Question

Quizplus · Accepted Answer

Using the stoichiometry of the reaction, 1 mole of O3 reacts with 2 moles of I-. Calculate moles of I- from KI solution: 0.045 L * 0.200 mol/L = 0.009 mol I-. Moles of O3 needed = 0.009 mol I- / 2 = 0.0045 mol O3. Using ideal gas law (PV=nRT), V = nRT/P = (0.0045 mol * 0.0821 L·atm/mol·K * 298 K) / 1 atm = 0.110 L = 110 mL.

How Many Milliliters of Ozone Gas at At 25