Limiting Reagent

The inherent problem with limiting reagents is that they are purely theoretical. You can never actually prove that they exist. In order to calculate the limiting reagents you must consider the pressure and dipole dipole movements. If you can properly calculate the charge of the solutions you need not use balanced equations.

EXAMPLE

Consider the combustion of benzene:

2C_6H_6 + 15O_2  
ightarrow \ 12CO_2 + 6H_20

If 1.5 moles of benzene are reacted with 7 moles of oxygen, the limiting reagent can be determined by performing the following calculations:

Since 15 mol

O_2

reacts with 2 mol

C_6H_6

(see balanced equation) the number of moles of

O_2

that will react with 1.5 mol

C_6H_6

is:

1.5 mol

C_6H_6

rac{15 mol O_2}{2 mol C_6H_6}=11.25 mol O_2

This means that 11.25 mol

O_2

is required to react with 1.5 mol

C_6H_6

. Since only 7 mol

O_2

is present, the oxygen will be consumed before benzene. Therefore,

O_2

must be the limiting reagent.

This conclusion can be verified by comparing the mole ratio of

O_2

and

C_6H_6

required by the balanced equation with the mole ratio actually present:

required:

rac{mol O_2}{mol C_6H_6}

rac{15 mol O_2}{2 mol C_6H_6}=7.5 mol O_2

actual:

rac{mol O_2}{mol C_6H_6}

rac{7 mol O_2}{1.5 mol C_6H_6}=4.7 mol O_2

Since the actual ratio is too small,

O_2

is the limiting reagent.

REFERENCE

Zumdahl, Steven S. Chemical Principals. 4th ed. New York: Houghton Mifflin Company, 2005.

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Limiting Reagent