Iodine pentafluoride is an incendiary agent, a substance that ignites combustible materials. This compound is usually made by passing fluorine gas over solid iodine, but it also forms when iodine monofluoride changes into the element iodine and iodine pentafluoride.
(a) Write a balanced equation, without including states, for the conversion of iodine monofluoride into iodine and iodine pentafluoride. (Use the lowest possible whole number coefficients.)
This looks straight forward to me.
5IF ==> 2I2 + IF5
You balance it by one of the redox methods you've studied. I changes from +1 on the left to 0 (for I2) on the right and +5(for IF5) on the right.
That is what I thought was the answer. I am taking classes online and each time I put the equation in the chemPad it tells me "Your answer contains an ambiguous or incomplete reaction equation. Check all the components on the product-side of the equation."
I don't know about this. From the answer on your key it appears that something is missing on the product side; however, the problem tells you iodine and IF5 are formed so surely there isn't anything missing. I googled iodine and came up with a link to Wikipedia. If you scroll down to "structure and bonding" and the last sentence in that section. it talks about the VERY weak I-I bond in I2 and it's propensity to dissociate into atomic I. So I might suggest that you try changing it from
5IF ==> 2I2 + IF5 to
5IF ==> 4I + IF5
Here is the link.
http://en.wikipedia.org/wiki/Iodine
Other than that I don't have any "far out" ideas. I would be very interested in knowing if that satisfies the key.
To write a balanced equation for the conversion of iodine monofluoride into iodine and iodine pentafluoride, we need to ensure that the number of atoms on both sides of the equation is the same.
The reactant is iodine monofluoride (IF), and the products are iodine (I2) and iodine pentafluoride (IF5).
The balanced equation is:
2 IF → I2 + IF5