# Equation for reaction of tin(IV) 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 Aug 21 '15 at 13:05

The procedure is correct, be careful with this:

1. The amount of iodine is 0.050, but the molecular amount ($I_{2}$) is 0,025.

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

$$\frac{\mathrm{Moles~of~Sn}}{\mathrm{Moles~of~I_2}} = \frac{0.017}{0.025} \approx \frac{2}{3}$$

1. Don't confuse! This equation is not well balanced. Stoichiometry is inherent of the reaction and is based on Mass Conservation Law; amount of substance tells you which reactive is limiting and which one is in excess.

$$$$\ce{Sn + 2I2 \leftrightharpoons SnI4}\tag{The reaction}$$\\ 0.017 + 0.025 = \,?$$

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

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