Limiting Reagent
Assume the reaction
A, B, C, and D are the chemical species, and a, b, c, and d are the coefficients of the balanced reaction. We can calculate
the masses corresponding to a and b in order to determine
what mass values of A and B to mix together.
We may have reasons, though, to use a mass of B less than what
is calculated by stoichiometry. Granted, at first this
sounds illogical. After all, it means that B will be used
up before all of A has a chance to react, thus, part of A
will be wasted. If we decide reduce the mass of B used, to
less than the stoichiometrically calculated quantity, we
say that B is the limiting reagent. There are
reasons for using a limiting reagent.
For the purpose of our discussion, we will give the designation
B to the limiting reagent. This is arbitrary. On a test, chances are you will be asked determine which reagent is the limiting reagent.
How to solve a complex limiting reagent problem the
problem will be explained here and then an example will be presented:
- You may need to balance the equation.
- Convert the mass units given in the problem to moles
- Divide the stoichiometric coefficients of the balanced equation by the smallest coefficent so that the smallest
coefficient becomes one.
- Divide the moles of the masses given by the coefficient
corresponding to the smallest coefficient in the balanced
equation.
- Compare the two sets of mole ratios (1- those of the balanced equation, 2- those determined from the given masses).
example:
Chromium metal, Cr(s) can be prepared from reacting chromium oxide, Cr2O3(s), with aluminum, Al(s).
Al2O3(s) is a by-product. Assume a reaction of 15.000 g of chromium oxide with 10.000 g of aluminum
metal.
Cr2O3(s) + Al(s) --> Cr(s) + Al2O3(s)
The above reaction is not balanced. It is balanced below:
2 Cr2O3(s) + 4 Al(s) --> 4 Cr(s) + 2 Al2O3(s)
We must now calculate the molecular weight of chromium oxide, and use the atomic weight of aluminum given on the periodic table (26.98 g).
2 Cr = 2 (52.00) = 104.00
3 O = 3 (16.00) = 48.00
------
152.00
You can use Chemfinder to look up information about a chemical such as its molecular weight.
We need the smallest reactant coefficient set to one. This can be achieved by dividing each coefficient by 2:
1 Cr2O3(s) + 2 Al(s) --> 2 Cr(s) + 1 Al2O3(s)
Now, to convert the given masses to moles:
mole Cr2O3
15.000 g Cr2O3 * -------------- = 0.098684 moles Cr2O3
152.00 g Cr2O3
mole Al
10.000 g Al * ---------- = 0.3706 moles Al
26.98 g Al
Since Cr2O3 is set to one, we divide each coefficient
above by 0.098684:
0.098684 0.3706
-------- = 1.00000 -------- = 3.755
0.098684 0.098684
We now compare the ratios:
Coefficients of balanced reaction:
Moles from given masses:
From the numbers, there is too much Aluminum present, thus
chromium oxide is the limiting reagent.
Reasons for using a limiting reagent- Assume that B is the limiting reagent.
- B is a very expensive chemical, and A is very inexpensive. If we use stoichiometric quantities, then the reaction will not go to 100% conversion. However, by using excess A (it is cheap!), we guarantee that all B will react, thus increasing the profitability of the synthesis.
- Excess A is used because at the high temperatures used for reaction, A also undergoes a side reaction to something else, and is thus wasted.
- A is volatile, so some is lost during the reaction before it has a chance to react.
- Perhaps the A+B reaction is slow, and A is very inexpensive, and by increasing the concentration of A, less time is required.
Perhaps the system is set up so that the excess A is recovered
and recycled. Chemical engineers are paid good money to be clever.
Note: When you discuss Le'Chatlier's Principle, you will cover
ideas used in the above discussion.
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Last Revised 01/25/98.
Copyright ©1998 by William L. Dechent. All rights reserved.