The chemical equation below shows the reaction between tin (sn) and hydrogen fluoride (hf).

sn + 2hf mc017-1.jpg snf2 + h2

the molar mass of hf is 20.01 g/mol. how many moles of sn are required to react with 40 g of hf?

This is a grams to moles stoichiometry problem. Lets give it a try! :)

The first step is ALWAYS going to be our given over one. Our given is what we know, which is the 40 grams of HF, or, hydroflouric acid.

So our first step looks like this:

40g HF

 1

Next, we need to cancel out the grams of HF. By doing this, we will place the grams of HF on the BOTTOM of the 2nd step, so it cancels out. This particular mass will be the molar mass, which is 20.01. On top will be MOLES of HF, which is 1. Although it says 2 moles in our balanced equation, when we calculate mass, the moles in that step is always one. So far, our equation looks like this:

40g HF  1mol HF
x
    1         20.01g HF

Our third and final step is to cancel out the moles of HF to find our moles of Sn. For this step we will use our balanced equation. The balanced equation shows we have 2 moles of HF and 1 of Sn. So, we would place the HF on the bottom, and Sn on the top as that is what we are trying to find. Our full equation looks like this:

40g HF   1mol HF          1mol Sn
x x
     1         20.01g HF      2mol HF

To find our answer, we will multiply all of our top numbers together, which is 40. Then, we will multiply all of our bottom numbers together, which is 40.02. Then, we will divide our top total by our bottom total.

40

40.02 

This gives us 0.999.

So, 0.999 moles of Sn, Tin, are produced from 40 grams of Hydroflouric Acid.
:)

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