Balance the reaction below in acidic solution. IO₃^- + I^- $\rightarrow$ I₂
What volume of 0.352 M HCl is needed to produce 2.48 *10^-³ mol of iodine, I₂, with an excess of KIO₃ and KI?

Question

Balance the reaction below in acidic solution. IO₃^- + I^- $\rightarrow$ I₂ What volume of 0.352 M HCl is needed to produce 2.48 *10^-³ mol of iodine, I₂, with an excess of KIO₃ and KI?

Quizplus · Accepted Answer

The balanced reaction in acidic solution is:IO₃^- + 5I^- + 6H⁺ → 3I₂ + 3H₂OTo produce 2.48 × 10^-3 mol of I₂, we need 5 × 2.48 × 10^-3 = 1.24 × 10^-2 mol of I^-.The volume of 0.352 M HCl needed to provide 1.24 × 10^-2 mol of H⁺ is:V = n/M = (1.24 × 10^-2 mol) / (0.352 M) = 0.0352 L = 35.2 mLTherefore, the correct answer is A) 14.1 mL.

Balance the Reaction Below in Acidic Solution →\rightarrow→ I2 What Volume of 0

Balance the Reaction Below in Acidic Solution $\rightarrow$ I₂
What Volume of 0