CHEM 112 Lecture Notes - Lecture 9: Limiting Reagent, Reagent

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11 Nov 2017
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Limiting reagents
I’e leaed a ey siple ethod fo liitig eaget uestios hih ill e
very easy for you to remember. Just like the empirical formula method, we use
a table to hold all our information.
For example;
The reaction of 167 g Fe2O3 with 85.8 g CO produces 72.3 g Fe.
Determine the limiting reactant, theoretical yield, and percent yield.
The thing to remember with limiting reagents is that the reactants always
determine the products so you always use the masses and information of the
reactants to find the amount of product.
The first thing you need to do is rewrite the equation and write the
information you are given in the question above the respective substances. Do
this so you can see the n=m/M equation clearer.
167g 85.8g 72.3g
Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g)
n= 0.9537 mol 3.0632 mol
We’ll o ostut a tale to put all ou ifoatio i:
Fe2O3
3 CO
2 Fe
3 CO2
Initial mols
0.9537 mol
3.0632 mol
2.0421 mol
Reacting mols
End mols
We use a method called hop, skip, ad jup to find the moles of product
that each reagent/reactant produces. It is as follows:
Hop = divide
Skip = multiply
Jump = to the answer
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