# 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?

Nov 29, 2016

You need a stoichiometric equation. You can make a mass of approx. $50 \cdot g$

#### Explanation:

$\frac{1}{2} {I}_{2} \left(g\right) + \frac{7}{2} {F}_{2} \left(g\right) \rightarrow I {F}_{7}$

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.

$\text{Moles of difluorine}$ $=$ $\frac{25.0 \cdot g}{2 \times 19 \cdot g \cdot m o {l}^{-} 1} = 0.658 \cdot m o l$.

Given stoichiometric iodine, we can form $\frac{2}{7}$ equiv of the interhalogen, i.e. $\frac{2}{7} \times 0.658 \cdot m o l = 0.188 \cdot m o l$.

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.