What mass of iodine is produced when 7.1 g of chlorine reacts with potassium iodide?

Dec 27, 2016

We need a stoichiometric equation.............and predict that approx. $25 \cdot g$ of ${I}_{2}$ will result.............

Explanation:

$C {l}_{2} \left(a q\right) + 2 K I \left(a q\right) \rightarrow {I}_{2} \left(s\right) + 2 K C l \left(a q\right)$

The balanced equation is crucial. It tells you that one equiv of diiodine results from one equiv of dichlorine. How did you know that this reaction is valid? Well, you simply have to learn it. Fluorine is a stronger oxidant than chlorine (in fact fluorine is a stronger oxidant than anything). Chlorine is a stronger oxidant than bromine. And bromine is a stronger oxidant than iodine.

Why is the halogen lower on the table a stronger oxidant? Because the valence electron of the lower halogen is in a higher valence shell, and this electron tends to be donated to the halogen above it, thereby reducing the upper halogen, and OXIDIZING the starting halogen.

TO illustrate this we could write the redox half equation:

$\frac{1}{2} C {l}_{2} \left(a q\right) + {e}^{-} \rightarrow C {l}^{-} \left(a q\right)$

And of course the iodide anion is the electron source, the reductant:

${I}^{-} \rightarrow \frac{1}{2} {I}_{2} \left(s\right) + {e}^{-}$

To finally get to your question, we have a molar quantity of $\frac{7.1 \cdot g}{70.9 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.100 \cdot m o l$ of dichlorine gas. And given the equation, we get a stoichiometric quantity of ${I}_{2}$, i.e. $2 \times 126.9 \cdot g \cdot m o {l}^{-} 1 \times 0.100 \cdot m o l \cong 25 \cdot g$