\textbf{85 --} According to the given equation, if $3.95$ grams of $\mathrm{KMnO_4}$ reacts with a sufficient amount of hydroiodic acid solution and $12.7$ grams of a two-atom molecule is formed, what is the percent yield of the reaction? $(\mathrm{O = 16,\ K = 39,\ Mn = 55,\ I = 127 : g.mol^{-1}})$
$$\mathrm{KMnO_4(s) + HI(aq) \rightarrow MnI_2(aq) + H_2O(l) + KI(aq) + I_2(s)}$$
(1) 75 \hfill (2) 80 \hfill (3) 85 \hfill (4) 90
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