Equation for reaction of tin and iodine to produce tin(IV) iodide Tin(IV) iodide is prepared by direct combination of the elements.
Add 2.00 g of granulated tin to a solution of 6.35 g of iodine. Write an equation for this reaction.

Attempt:
1. Determine the amount of tin using moles = mass/molar mass. This yields a result of 0.017 mol $\ce{Sn}$.
2. Determine the amount of iodine using moles = mass/molar mass. This yields a result of 0.5 mol $\ce{I}$.
3. Determine the ratio of tin to iodine. Using relative amounts, for every 1 mol $\ce{Sn}$ there are 3 mol $\ce{I}$, so the ratio is 1:3. Because iodine is a diatomic molecule, I will adjust accordingly, so the ratio becomes 2 mol $\ce{Sn}$ to 3 mol $\ce{I2}$ or 2:6.
4. Construct the formula:
$$\ce{2Sn + 3I2 -&gt; SnI4}$$
with excess tin and iodine.
Is my procedure correct?
The balanced chemical equation is independent of the amounts of each reactant added. In this case we can easily deduce the equation to be: $$\ce{Sn + 2I2 -&gt; SnI4}$$

What the given masses tell us is that there is excess tin being added. We can see this because the $\ce{Sn:I2}$ ratio added is $1:1.5$ whereas the stoichiometric ratio is $1:2$ and so there is less iodine than is required to completely react with the added tin - or in other words there is an excess of tin.