Question #44671

A $2.29 \text{ g}$ sample of a pure oxide of manganese, \ce{Mn_xO_y}, is fused with excess \ce{KOH} in the presence of \ce{O2} gas. The oxide is completely converted to a green melt, consuming exactly $560 \text{ mL}$ of \ce{O2} gas at STP ($22.4 \text{ L/mol}$).

The green melt is extracted with water and acidified with dilute \ce{H2SO4}, causing complete disproportionation of the green species into a black precipitate and a purple solution.

After filtering the black precipitate, the purple filtrate is adjusted to a faintly alkaline pH ($pH \approx 8$) and treated with excess \ce{KI}. This results in the complete reduction of the purple species, forming more of the black precipitate and a soluble iodine-containing species.

Identify the empirical formula of the original oxide \ce{Mn_xO_y} and calculate the exact number of moles of the soluble iodine-containing species formed.

(Assume atomic masses: Mn = $55 \text{ u}$, O = $16 \text{ u}$)