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# Limiting reagent

In chemistry, the limiting reagent, or also called the "limiting reactant", is the chemical that determines how far the reaction will go before the chemical in question gets used up, causing the reaction to stop. The chemical of which there are fewer mols than the proportion requires is the limiting reagent.

## Example

Consider the combustion of benzene:

1.5 mol C6H6 × $\frac{15 mol O_2}{2 mol C_6H_6}=11.25 mol O_2$

This means that 11.25 mol O2 is required to react with 1.5 mol C6H6. Since only 7 mol O2 is present, the oxygen will be consumed before benzene. Therefore, O2 must be the limiting reagent.

This conclusion can be verified by comparing the mole ratio of O2 and C6H6 required by the balanced equation with the mole ratio actually present:

required: $\frac{mol O_2}{mol C_6H_6}$ = $\frac{15 mol O_2}{2 mol C_6H_6}=7.5 mol O_2$

actual: $\frac{mol O_2}{mol C_6H_6}$ = $\frac{7 mol O_2}{1.5 mol C_6H_6}=4.7 mol O_2$

Since the actual ratio is too small, O2 is the limiting reagent.

Consider a typical thermite reaction:

If 20.0 g of Fe2O3 are reacted with 8.00 g Al(s) in the thermite reaction, Which reactant is limiting?. $Fe_2O_3(s) + 2 Al(s) \rightarrow 2 Fe(l) + Al_2O_3(s)\,$

First, determine how many moles of Fe(l) can be produced from either reactant.

Moles produced of Fe from reactant Fe2O3 $mol Fe_2O_3 = \frac{grams Fe_2O_3}{g/mol Fe_2O_3}\,$ $mol Fe_2O_3 = \frac{20.0 g}{159.7 g/mol} = 0.127 mol\,$ $mol Fe = (2)(0.127) = 0.254 mol Fe\,$

Moles produced of Fe from reactant Al $mol Al = \frac{grams Al}{g/mol Al}\,$ $mol Al = \frac{8.00 g}{26.98 g/mol} = 0.297 mol\,$ $mol Fe = \frac{(2)(0.297)}{2} = 0.297 mol Fe\,$

Because the moles Fe produced from Fe2O3(0.254mol) is less than the moles Fe produced from Al(0.297mol), Fe2O3 is the limiting reagent.

By looking at chemical equation for the thermite reaction, the limiting reagent can be found based on the ratio of moles of one reactant to another and the total atomic mass of the reactant compounds.

## Limiting reagent formula

There is a much simpler formula which can be used. However, you must first calculate the moles of both of the reagents in the reaction. Once the number of moles have been figured out, just simply fill in this equation (Reagent 1 being the first reactant, 2 being the second): $\mbox{Moles of Reagent 2} \times \frac{\mbox{Coefficient of Reagent 1}}{\mbox{Coefficient of Reagent 2}} - \mbox{Moles of Reagent 1}$

When the answer to the formula is less than zero, reagent 1 is the excess reagent. When the answer is larger than zero, reagent 1 is the limiting reagent. The number shows how much in excess one reagent is from another. If the answer for the formula is zero, both reagents are perfectly balanced. The unit of an answer is in moles.

## References

• Zumdahl, Steven S. Chemical Principals. 4th ed. New York: Houghton Mifflin Company, 2005. ISBN 0618372067.