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 0.400 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)

Question

Quizplus · Accepted Answer

The balanced chemical equation shows that 1 mole of O3 reacts with 2 moles of I-. The number of moles of I- in 45.00 mL of a 0.400 M aqueous solution of KI is:(0.400 mol/L) x (45.00 mL/1000 mL/L) = 0.0180 mol I-Therefore, the number of moles of O3 needed to react with 0.0180 mol of I- is:0.0180 mol I- x (1 mol O3 / 2 mol I-) = 0.00900 mol O3The volume of ozone gas at 25.0°C and 1.00 atm pressure needed to contain 0.00900 mol of O3 is:(0.00900 mol O3) x (22.4 L/mol O3) x (1000 mL/L) = 202 mL O3Therefore, the correct answer is A) 220 mL.

How Many Milliliters of Ozone Gas at At 25