Question

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

Answer 1

Step-by-step explanation:

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.

`IF_(7) + 6H_(2)O \rightarrow H_(5)IO_(6) + 7HF`

As it is given that there are
`3.72 * 10^(-2)` moles of iodine heptafluoride are present. Molar mass of
`IF_(7)` is 259.9 g/mol. Molar mass of
`HIO_(4)` is 227.94 g/mol and molar mass of HF is 20.01 g/mol.

Now, according to the reaction equation
`3.72 * 10^(-2)` M
`IF_(7)` gives
`3.72 * 10^(-2)` M
`H_(5)IO_(6)`.

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

Hence, calculate the concentration of
`H_(5)IO_(6)` into the solution as follows.

Concentration =
`\frac{\text{no. of moles}}{volume}`

=
`(3.72 * 10^(-2))/(0.795 L)`

=
`4.7 * 10^(-2)` mol/L

or, = 0.047 M

Now, as 1 mol of
`IF_(7)` produces 7 mol HF. So,
`7 * 3.72 * 10^(-2) = 26.04 * 10^(-2)` M HF.

Therefore, concentration of HF will be calculated as follows.

Concentration of HF =
`(0.2604)/(0.795)`

= 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.