Question

Hg5 (IO6 )2 oxidizes KI to I2 in acid medium and the other product containing iodine is K2HgI4 . If the I2 liberated in the reaction requires 0.004 mole of Na2S2O3 , the number of moles of Hg5 (IO6 )2 that have reacted is : ​​(A) 10^-3 (B) 10^-4 (C) 2.5 × 10^-4 (D) 2.5 × 10^-2

Answer 1

Answer:

The correct answer is option C.

Explanation:

$2Na_2S_2O_3 + I_2\rightarrow Na_2S_4O_6 + 2NaI$

Sodium thiosulfate react with iodine to produce tetrathionate sodium and sodium iodide.

According to reaction. 2 moles of sodium thiosulfate reacts with 1 mole of iodine gas.

Then 0.004 moles of Sodium thiosulfate will react with:

$\frac{1}{2}\times 0.004 mol=0.002 mol$ iodine gas

Moles of iodine reacted with sodium thiosulfate is equal to the moles of iodine gas liberated from the given reaction:

$Hg_5(IO_6)_2+34KI+24HCl\rightarrow 8I_2+5K_2HgI_4+12H_2O+24KCl$

According to reaction, 1 mol of $Hg_5(IO_6)_2$ gives 8 moles of iodine gas.

Then 0.002 mol of iodine gas will be obtained from:

$\frac{1}{8}\times 0.002 ml=2.5\times 10^{-4} mol$ of  $Hg_5(IO_6)_2$.

Hence, the correct answer is option C.