The reactant that runs out first controls how much product forms

Definition

Limiting reactant

The reactant that is completely used up in a reaction, and which therefore controls the maximum mass of product formed.

Reactants are rarely mixed in exactly the ratio the equation calls for.
One of them is used up before the other, and the reaction stops at that point.
Whatever is left over is described as being in excess.
The amount of product is set entirely by the reactant that ran out.
Adding more of the excess reactant changes nothing, because there is nothing left for it to react with.

Key Idea

The limiting reactant caps the product, whatever else is in the flask.
The excess reactant is left behind, so it appears in neither the product mass nor the calculation of it.

Comparing amounts, not masses, identifies the limiting reactant

Definition

Mole

The amount of a substance that contains the Avogadro constant of particles, and which has a mass in grams equal to its relative particle mass.

Equal masses of two substances are not equal amounts, because their particles have different masses.
Convert each reactant's mass into an amount first: $\frac{m}{M_r}$
Divide each amount by that substance's coefficient in the balanced equation.
The smallest of those results identifies the limiting reactant.
A reactant present in the larger mass can still be the limiting one.

Common Mistake

Masses cannot be compared directly, because a gram of magnesium is a different number of particles from a gram of oxygen.
The coefficient has to be divided out, because two moles of magnesium react with only one mole of oxygen.

A worked limiting reactant calculation

Burn $g12.0\ \text{g}$ of magnesium with $g10.0\ \text{g}$ of oxygen, where the equation is $2Mg+O2→2MgO2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ .
Convert both masses, using $Ar(Mg)=24A_r(\text{Mg}) = 24$ and $Mr(O2)=32M_r(\text{O}_2) = 32$ :
$moln(\text{Mg}) = \frac{12.0}{24} = 0.500\ \text{mol}$ $moln(\text{O}_2) = \frac{10.0}{32} = 0.3125\ \text{mol}$
Divide each amount by its coefficient: magnesium gives $0.500 \div 2 = 0.250$ , and oxygen gives $0.3125 \div 1 = 0.3125$ .
Magnesium gives the smaller result, so magnesium is the limiting reactant and the oxygen is in excess.
The reaction consumes only $mol0.250\ \text{mol}$ of oxygen, leaving $0.0625\ \text{mol}$ unreacted.
The ratio $2Mg:2MgO2\text{Mg} : 2\text{MgO}$ means that $mol0.500\ \text{mol}$ of magnesium gives $mol0.500\ \text{mol}$ of magnesium oxide.
With $Mr(MgO)=40M_r(\text{MgO}) = 40$ , the mass of product is $0.500 \times 40 = 20.0\ \text{g}$ .

Note

The divided figures are the comparison, $0.250$ for magnesium against $0.3125$ for oxygen.
The excess amount answers a different question, about what is left over rather than what is made.

Deducing the stoichiometry from measured masses

Definition

Stoichiometry

The ratio in which substances react and are produced, given by the balancing numbers in a balanced chemical equation.

Measured masses of the reactants and products can be turned back into the balancing numbers.
Divide each substance's mass by its relative formula mass, which gives a set of comparison numbers.
Divide every one of those by the smallest, then scale the results to whole numbers.
Those whole numbers are the coefficients in the balanced equation.
For $g4.8\ \text{g}$ of magnesium, $g3.2\ \text{g}$ of oxygen and $g8.0\ \text{g}$ of magnesium oxide, the comparison numbers are $0.200$ , $0.100$ and $0.200$ .
Dividing by $0.100$ gives $2$ , $1$ and $2$ , so the equation is $2Mg+O2→2MgO2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ .
The masses check out as well, because $4.8 + 3.2 = 8.0$ .
At Higher tier those comparison numbers are the amounts in moles, and the method is identical.

Common Mistake

The masses themselves are never the ratio, because the ratio compares amounts rather than masses.
Each comparison number uses the right relative mass, so oxygen is divided by $32$ for $O2\text{O}_2$ and not by $16$ .

Checking the finished equation

Count each kind of atom on both sides of the equation you have deduced.
Confirm the coefficients are whole numbers with no common factor left in them.
Confirm the total mass of the reactants matches the total mass of the products.
When a calculation follows, start from the limiting reactant rather than from whichever mass came first.
Check the final unit, since an amount is in $mol\text{mol}$ and a mass in $g\text{g}$ .

Exam technique

The reason for choosing a limiting reactant is the divided figure, never which mass or amount happens to be larger.
The product mass comes from the limiting reactant alone, so the excess never enters that calculation.
Masses are converted before any ratio is taken, because the coefficients compare amounts.

Self review

What is meant by the limiting reactant?
Why does adding more of the excess reactant not increase the mass of product?
How do you decide which of two reactants is the limiting one?
Why must the masses be converted before the ratio is taken?
What equation do $g4.8\ \text{g}$ of magnesium, $g3.2\ \text{g}$ of oxygen and $g8.0\ \text{g}$ of magnesium oxide give?

The reactant that runs out first controls how much product forms

Definition

Limiting reactant

The reactant that is completely used up in a reaction, and which therefore controls the maximum mass of product formed.

Reactants are rarely mixed in exactly the ratio the equation calls for.
One of them is used up before the other, and the reaction stops at that point.
Whatever is left over is described as being in excess.
The amount of product is set entirely by the reactant that ran out.
Adding more of the excess reactant changes nothing, because there is nothing left for it to react with.

Key Idea

The limiting reactant caps the product, whatever else is in the flask.
The excess reactant is left behind, so it appears in neither the product mass nor the calculation of it.

Comparing amounts, not masses, identifies the limiting reactant

Definition

Mole

The amount of a substance that contains the Avogadro constant of particles, and which has a mass in grams equal to its relative particle mass.

Equal masses of two substances are not equal amounts, because their particles have different masses.
Convert each reactant's mass into an amount first: $\frac{m}{M_r}$
Divide each amount by that substance's coefficient in the balanced equation.
The smallest of those results identifies the limiting reactant.
A reactant present in the larger mass can still be the limiting one.

Common Mistake

Masses cannot be compared directly, because a gram of magnesium is a different number of particles from a gram of oxygen.
The coefficient has to be divided out, because two moles of magnesium react with only one mole of oxygen.

A worked limiting reactant calculation

Burn $g12.0\ \text{g}$ of magnesium with $g10.0\ \text{g}$ of oxygen, where the equation is $2Mg+O2→2MgO2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ .
Convert both masses, using $Ar(Mg)=24A_r(\text{Mg}) = 24$ and $Mr(O2)=32M_r(\text{O}_2) = 32$ :
$moln(\text{Mg}) = \frac{12.0}{24} = 0.500\ \text{mol}$ $moln(\text{O}_2) = \frac{10.0}{32} = 0.3125\ \text{mol}$
Divide each amount by its coefficient: magnesium gives $0.500 \div 2 = 0.250$ , and oxygen gives $0.3125 \div 1 = 0.3125$ .
Magnesium gives the smaller result, so magnesium is the limiting reactant and the oxygen is in excess.
The reaction consumes only $mol0.250\ \text{mol}$ of oxygen, leaving $0.0625\ \text{mol}$ unreacted.
The ratio $2Mg:2MgO2\text{Mg} : 2\text{MgO}$ means that $mol0.500\ \text{mol}$ of magnesium gives $mol0.500\ \text{mol}$ of magnesium oxide.
With $Mr(MgO)=40M_r(\text{MgO}) = 40$ , the mass of product is $0.500 \times 40 = 20.0\ \text{g}$ .

Note

The divided figures are the comparison, $0.250$ for magnesium against $0.3125$ for oxygen.
The excess amount answers a different question, about what is left over rather than what is made.

Deducing the stoichiometry from measured masses

Definition

Stoichiometry

The ratio in which substances react and are produced, given by the balancing numbers in a balanced chemical equation.

Measured masses of the reactants and products can be turned back into the balancing numbers.
Divide each substance's mass by its relative formula mass, which gives a set of comparison numbers.
Divide every one of those by the smallest, then scale the results to whole numbers.
Those whole numbers are the coefficients in the balanced equation.
For $g4.8\ \text{g}$ of magnesium, $g3.2\ \text{g}$ of oxygen and $g8.0\ \text{g}$ of magnesium oxide, the comparison numbers are $0.200$ , $0.100$ and $0.200$ .
Dividing by $0.100$ gives $2$ , $1$ and $2$ , so the equation is $2Mg+O2→2MgO2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ .
The masses check out as well, because $4.8 + 3.2 = 8.0$ .
At Higher tier those comparison numbers are the amounts in moles, and the method is identical.

Common Mistake

The masses themselves are never the ratio, because the ratio compares amounts rather than masses.
Each comparison number uses the right relative mass, so oxygen is divided by $32$ for $O2\text{O}_2$ and not by $16$ .

Checking the finished equation

Count each kind of atom on both sides of the equation you have deduced.
Confirm the coefficients are whole numbers with no common factor left in them.
Confirm the total mass of the reactants matches the total mass of the products.
When a calculation follows, start from the limiting reactant rather than from whichever mass came first.
Check the final unit, since an amount is in $mol\text{mol}$ and a mass in $g\text{g}$ .

Exam technique

The reason for choosing a limiting reactant is the divided figure, never which mass or amount happens to be larger.
The product mass comes from the limiting reactant alone, so the excess never enters that calculation.
Masses are converted before any ratio is taken, because the coefficients compare amounts.

Self review

What is meant by the limiting reactant?
Why does adding more of the excess reactant not increase the mass of product?
How do you decide which of two reactants is the limiting one?
Why must the masses be converted before the ratio is taken?
What equation do $g4.8\ \text{g}$ of magnesium, $g3.2\ \text{g}$ of oxygen and $g8.0\ \text{g}$ of magnesium oxide give?

2.6.6 Limiting reactants and stoichiometry

The reactant that runs out first controls how much product forms

Comparing amounts, not masses, identifies the limiting reactant

A worked limiting reactant calculation

Deducing the stoichiometry from measured masses

Checking the finished equation

Review 2.6.6 Limiting reactants and stoichiometry by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

2.6.6 Limiting reactants and stoichiometry

The reactant that runs out first controls how much product forms

Comparing amounts, not masses, identifies the limiting reactant

A worked limiting reactant calculation

Deducing the stoichiometry from measured masses

Checking the finished equation

Review 2.6.6 Limiting reactants and stoichiometry by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

2.6.6 Limiting reactants and stoichiometry

The reactant that runs out first controls how much product forms

Comparing amounts, not masses, identifies the limiting reactant

A worked limiting reactant calculation

Deducing the stoichiometry from measured masses

Checking the finished equation

Review 2.6.6 Limiting reactants and stoichiometry by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

2.6.6 Limiting reactants and stoichiometry Revision Guide

2.6.6 Limiting reactants and stoichiometry

The reactant that runs out first controls how much product forms

Comparing amounts, not masses, identifies the limiting reactant

A worked limiting reactant calculation

Deducing the stoichiometry from measured masses

Checking the finished equation

Review 2.6.6 Limiting reactants and stoichiometry by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

2.6.6 Limiting reactants and stoichiometry Revision Guide