During a chemical reaction, there is not always the correct amount of each reactant present in the ratio in the balanced equation. Obviously, more = excess and less = limiting
These questions take a while to do and require a lot of steps so remember to keep organized. Alright, here is how to do it!
Lets start off with an example:
Fe + S --> FeS
A sample of 111.6 g of Fe is mixed with 96.3 g of S.
First thing to check is that the equation is balanced. After that, you must calculate the moles present.
moles present Fe = 111.6 g Fe x 1 mol Fe/55.8 g Fe = 2.00 mol present Fe
moles present S = 96.3 g S x 1 mol S/32.1 g S = 3.00 mol S present
So now we have our moles present. But how do we know which one is limiting and which one is excess? Well, we must calculate the moles needed.
Use the coefficients in the equation to determine the moles needed.
moles needed S = 2.00 mol Fe x 1 mol S / 1 mol Fe = 2.00 mol S needed
Using moles present and moles needed we can calcullate the excess and limiting reactants.
With 3.00 mols S present, and 2.00 mols S needed, S is the excess reactant. 3.00-2.00=1.00 g in excess
Therefore, if S is the excess reactant than Fe is the limiting reactant!
A question that will typically be asked is:
When this reaction is carried out, what mass of FeS will actually be produced?
To calculate the mass of FeS actually formed in this reaction, we have to start with the limiting reactant (this time Fe). We can't use the limiting reactant since not all of it reacts.
To get the answer, use the stoicheometry we have learned.
Remember to plan first.
mass Fe --> moles Fe ---> moles FeS --> mass FeS
Just write it all out and you are good to go! Lets try another.
4FeS + 7O2 --> 2Fe2O3 + 4SO2
moles present FeS = 439.5 g FeS x 1 mol FeS/87.9 g FeS = 5.00 mol FeS
moles present O2 = 256.0 g O2 x 1 mol O2/32.0 g O2 = 8.00 mol O2
moles needed O2 = 5.00 mol FeS x 7 mol O2/4 mol FeS = 8.75 mol O2
Therefore O2 is the limiting reactant and FeS is the excess reactant.
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