Add $2.0\:\mathrm g$ of tin to $6.35\:\mathrm g$ iodine.


  1. There are 0.017 moles of tin.

  2. There are 0.0500 moles of iodine. (0.0250 moles of $\ce{I2}$).

  3. Prepare ratio. For every 1.5 $\ce{I2}$s there are 1 $\ce{Sn}$. This doesn't make it easy, therefore there are 3 $\ce{I2}$s for every 2 $\ce{Sn}$.

  4. Construct equation. $\ce{2Sn + 3I2 -> 2SnI4}$ (doubled because we doubled the amount of reactants so that it would make it easier to understand).

But this means that there is too little iodine and no excess reactant! Where did I go wrong?!

  • 2
    $\begingroup$ Duplicate of chemistry.stackexchange.com/questions/35309/… $\endgroup$ – Todd Minehardt Sep 2 '15 at 15:22
  • $\begingroup$ @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). $\endgroup$ – user19586 Sep 2 '15 at 15:25
  • $\begingroup$ 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. $\endgroup$ – bon Sep 2 '15 at 16:23
  • $\begingroup$ @bon Okay, so the equation is Sn + 2I2 --> SnI4. But how do you conclude that there is excess tin? $\endgroup$ – user19586 Sep 2 '15 at 18:02
  • $\begingroup$ 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. $\endgroup$ – bon Sep 2 '15 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.


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