**Passage 12:** To mL of solution, mL of and excess acid were added. The liberated was then titrated — Redox Reactions and Volumetric Analysis Chemistry Question
Question
**Passage 12:** To $25$ mL of $0.1 M$ $KI$ solution, $25$ mL of $0.1 M$ $KIO_3$ and excess acid were added. The liberated $I_2$ was then titrated with $Na_2S_2O_3$ solution. <br> **Q1:** The moles of $I_2$ liberated from the reaction $IO_3^- + 5I^- + 6H^+ \rightarrow 3I_2 + 3H_2O$ is limited by:
Answer: A
💡 Solution & Explanation
**Step 1:** Moles $KI = 0.025 \times 0.1 = 0.0025$. <br>**Step 2:** Moles $KIO_3 = 0.025 \times 0.1 = 0.0025$. <br>**Step 3:** Required ratio $I^- : IO_3^-$ is $5:1$. <br>**Step 4:** For $0.0025$ moles $KIO_3$, we need $0.0125$ moles $KI$. Since we only have $0.0025$, $KI$ is the limiting reagent.
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