# 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 moles than the proportion requires is the limiting reagent.

## Example

Consider the combustion of benzene:

1.5 mol ${\displaystyle C_{6}H_{6}}$ × ${\displaystyle {\frac {15molO_{2}}{2molC_{6}H_{6}}}=11.25molO_{2}}$

This means that 11.25 mol ${\displaystyle O_{2}}$ is required to react with 1.5 mol ${\displaystyle C_{6}H_{6}}$. Since only 7 mol ${\displaystyle O_{2}}$ is present, the oxygen will be consumed before benzene. Therefore, ${\displaystyle O_{2}}$ must be the limiting reagent.

This conclusion can be verified by comparing the mole ratio of ${\displaystyle O_{2}}$ and ${\displaystyle C_{6}H_{6}}$ required by the balanced equation with the mole ratio actually present:

required: ${\displaystyle {\frac {molO_{2}}{molC_{6}H_{6}}}}$ = ${\displaystyle {\frac {15molO_{2}}{2molC_{6}H_{6}}}=7.5molO_{2}}$

actual: ${\displaystyle {\frac {molO_{2}}{molC_{6}H_{6}}}}$ = ${\displaystyle {\frac {7molO_{2}}{1.5molC_{6}H_{6}}}=4.7molO_{2}}$

Since the actual ratio is too small, ${\displaystyle O_{2}}$ 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?.

${\displaystyle Fe_{2}O_{3}(s)+2Al(s)\rightarrow 2Fe(l)+Al_{2}O_{3}(s)\,}$

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

Moles produced of ${\displaystyle Fe}$ from reactant ${\displaystyle Fe_{2}O_{3}}$

${\displaystyle molFe_{2}O_{3}={\frac {gramsFe_{2}O_{3}}{g/molFe_{2}O_{3}}}\,}$

${\displaystyle molFe_{2}O_{3}={\frac {20.0g}{159.7g/mol}}=0.127mol\,}$

${\displaystyle molFe=(2)(0.127)=0.254molFe\,}$

Moles produced of ${\displaystyle Fe}$ from reactant ${\displaystyle Al}$

${\displaystyle molAl={\frac {gramsAl}{g/molAl}}\,}$

${\displaystyle molAl={\frac {8.00g}{26.98g/mol}}=0.297mol\,}$

${\displaystyle molFe={\frac {(2)(0.297)}{2}}=0.297molFe\,}$

Because the moles ${\displaystyle Fe}$ produced from ${\displaystyle Fe_{2}O_{3}}$(${\displaystyle 0.254mol}$) is less than the moles ${\displaystyle Fe}$ produced from ${\displaystyle Al}$(${\displaystyle 0.297mol}$), ${\displaystyle Fe_{2}O_{3}}$ 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 and 2 being the second):

${\displaystyle {\mbox{Moles of Reagent 2}}\times {\frac {\mbox{Coefficient of Reagent 1}}{\mbox{Coefficient of Reagent 2}}}-{\mbox{Moles of Reagent 1}}}$

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.