13. Upon complete interaction of a solution containing 0.88 g of compound A and an equivalent amount of hydroiodic acid, only a suspension of iodine in water was formed. The iodine that precipitated reacted with an excess of sodium sulfide solution, resulting in a precipitate with a mass of 0.48 g. Determine the composition of compound A by calculation.
13. Since only iodine and water were formed, A could only contain iodine, hydrogen, and oxygen atoms. A is an oxidizer. Possible compounds - $\mathrm{H}_{2} \mathrm{O}_{2}$ or oxygen-containing iodine acids. $\mathrm{I}_{2}+\mathrm{Na}_{2} \mathrm{~S}=2 \mathrm{NaI}+$ S. Precipitate - sulfur. Let's calculate the number of moles of iodine formed in the reaction with $A$: $n\left(I_{2}\right)=0.48 / 32=0.015$ mol. Let's check the hypothesis that $\mathrm{A}=\mathrm{H}_{2} \mathrm{O}_{2}$

$2 \mathrm{HI}+\mathrm{H}_{2} \mathrm{O}_{2}=\mathrm{I}_{2}+2 \mathrm{H}_{2} \mathrm{O}$

$n\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)=n\left(I_{2}\right)=0.015$ mol; $\quad m\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)=0.512$ - does not correspond to the conditions. Let's write the general equation for the interaction of $HI$ with an oxygen-containing iodine acid - $(2 x-1) HI+\mathrm{HIO}_{x}=x I_{2}+x H_{2} \mathrm{O}$. According to the equation: $n\left(H_{0} O_{x}\right)=0.88 /\left(127+1+16^{*} x\right)=0.015 / x x=3$, this is $\mathrm{HIO}_{3}$