What volume (in mL) of 1.0 M is required to completely oxidize 16.8 g of ions to in a reaction where — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

Moles of $Fe^{2+} = 16.8 / 56 = 0.3$ mol. The n-factor of $Fe^{2+} \rightarrow Fe^{3+}$ is 1. Thus, equivalents of reductant = 0.3. The reduction half for the oxidant is $NO_3^- \rightarrow NO$. N changes from +5 to +2, so the n-factor is 3. By Law of Equivalence: $M \times n \times V(in L) = 0.3 \Rightarrow 1.0 \times 3 \times V = 0.3 \Rightarrow V = 0.1\ L = 100$ mL.