Question

What is the maximum amount of `IF_7` that can be obtained from 25.0 g fluorine in the equation `I_2 + F_2 -> IF_7`?

Answer 1

You need a stoichiometric equation. You can make a mass of approx. `50*g`

Explanation:

`1/2I_2(g) + 7/2F_2(g)rarr IF_7`

I introduced the half-coefficient because it makes the arithmetic a little bit easier. I am certainly free to do so. The stoichiometry dictates that 1 equiv of diiodine reacts with 7 squiv difluorine.

`"Moles of difluorine"` `=` `(25.0*g)/(2xx19*g*mol^-1)=0.658*mol`.

Given stoichiometric iodine, we can form `2/7` equiv of the interhalogen, i.e. `2/7xx0.658*mol=0.188*mol`.

And thus a mass of `0.188*molxx259.90*g*mol^-1=??`.

It would not be my choice to do this reaction. I am too fond of my 10 fingers, and 2 eyes.