The densevapor iodinee heptafluoride reacts rapidly with water to give a mixture of periodic acid and hydrofluoric acid .

Write a balanced chemical equation for this reaction. Then determine the concentration (in moles per liter) of each of the acids that result from the complete reaction of 3.72×10-2moles of iodine heptafluoride with enough water to give a solution volume of 795 mL.

periodic acid _ M

hydrofluoric acid -M

When iodine heptafluoride reacts rapidly with water to give a mixture of periodic acid and hydrofluoric acid. Reaction equation for the same is as follows.

[tex]IF_{7} + 6H_{2}O \rightarrow H_{5}IO_{6} + 7HF[/tex]

As it is given that there are [tex]3.72 \times 10^{-2}[/tex] moles of iodine heptafluoride are present. Molar mass of [tex]IF_{7}[/tex] is 259.9 g/mol. Molar mass of [tex]HIO_{4}[/tex] is 227.94 g/mol and molar mass of HF is 20.01 g/mol.

Now, according to the reaction equation [tex]3.72 \times 10^{-2}[/tex] M [tex]IF_{7}[/tex] gives [tex]3.72 \times 10^{-2}[/tex] M [tex]H_{5}IO_{6}[/tex].

Also, the volume is given as 795 ml or 0.795 L (as 1 mL = 0.001 L).

Hence, calculate the concentration of [tex]H_{5}IO_{6}[/tex] into the solution as follows.

Concentration = [tex]\frac{\text{no. of moles}}{volume}[/tex]

= [tex]\frac{3.72 \times 10^{-2}}{0.795 L}[/tex]

= [tex]4.7 \times 10^{-2}[/tex] mol/L

or, = 0.047 M

Now, as 1 mol of [tex]IF_{7}[/tex] produces 7 mol HF. So, [tex]7 \times 3.72 \times 10^{-2} = 26.04 \times 10^{-2}[/tex] M HF.

Therefore, concentration of HF will be calculated as follows.

Concentration of HF = [tex]\frac{0.2604}{0.795}[/tex]

= 0.33 M

Thus, we can conclude that concentration of periodic acid is 0.047 M and concentration of hydrofluoric acid is 0.33 M.