The number of moles of required to oxidize one mole of in acidic medium is: — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

**Step 1:** $n$-factor of $KMnO_4$ (acidic) = $5$ ($Mn^{+7} \rightarrow Mn^{+2}$). <br>**Step 2:** $n$-factor of $FeC_2O_4 \rightarrow Fe^{3+} + 2CO_2$. Iron loses 1 $e^-$ ($Fe^{2+} \rightarrow Fe^{3+}$) and Carbon loses 2 $e^-$ ($C_2^{+3} \rightarrow 2C^{+4}$). Total $n$-factor = $1 + 2 = 3$. <br>**Step 3:** Equating equivalents: $n_1 \times M_1 = n_2 \times M_2 \Rightarrow 5 \times (\text{moles } KMnO_4) = 3 \times (1 \text{ mole } FeC_2O_4)$. <br>Moles of $KMnO_4 = 3/5$. [1-34]