# 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:

$$\ce{2Sn + 3I2 -> SnI4}$$

with excess tin and iodine.

Is my procedure correct?

• I think you are overcomplicating it. Your equation doesn't balance so it can't be correct. Surely the correct answer is just $\ce{Sn + 2I2 -> SnI4}$.
– bon
Commented Aug 21, 2015 at 13:05
• @ToddMinehardt That was my question (I couldn't remember my account), but I don't feel like I received a satisfactory answer (I don't know if you remember, but I couldn't leave a comment for some reason). Commented Sep 2, 2015 at 15:25
• I still don't see why you are attempting to calculate the equation using the masses given. The balanced chemical equation is independent of the amount of each substance which is added and in this case is easily determined as $\ce{Sn + 2I2 -> SnI4}$. All the masses tell you is that excess tin is being added.
– bon
Commented Sep 2, 2015 at 16:23
• @bon Okay, so the equation is Sn + 2I2 --> SnI4. But how do you conclude that there is excess tin? Commented Sep 2, 2015 at 18:02
• As you stated, the ratio of tin to iodine added is 1:1.5 but the ratio of tin to iodine in the equation is 1:2 so there is excess tin.
– bon
Commented Sep 2, 2015 at 18:16

The balanced chemical equation is independent of the amounts of each reactant added. In this case we can easily deduce the equation to be: $$\ce{Sn + 2I2 -> SnI4}$$

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.

The procedure is correct, be careful with this:

1. The amount of iodine is $$\pu{0.050 mol}$$, but the molecular amount $$\ce{I2}$$ is $$\pu{0.025 mol}$$.

2. For the ratio use the amount of point 1 and 2:

$$\frac{\text{Amount of Sn}}{\text{Amount of }\ce{I2}} = \frac{\pu{0.017 mol}}{\pu{0.025 mol}} \approx \frac{2}{3}$$

1. Don't confuse! This equation is not well balanced. Stoichiometry is inherent of the reaction and is based on the mass conservation law; amount of substance tells you which reactant is limiting and which one is in excess.

$$\ce{\underset{\pu{0.017 mol}}{Sn} + \underset{\pu{0.025 mol}}{2I2} <=> \underset{\pu{? mol}}{SnI4}}$$

Now with this you can tell which one is on excess and for how much.