4.2 Limiting and Non-Limiting Reactants | Stoichiometry | Chemistry 11

In a chemical reaction, reactants are often not mixed in the exact stoichiometric ratios required by the balanced chemical equation. This means one reactant will be completely used up before the others.

The limiting reactant (or limiting reagent) is the reactant that is completely consumed first in a chemical reaction. The amount of product formed is determined by this reactant.

The non-limiting reactant (or excess reactant) is the reactant that is present in a quantity greater than what is required to react completely with the limiting reactant. Some of this reactant will be left over after the reaction is complete.

Figure 4.2.1: Limiting and excess reactants diagram

Identifying the Limiting Reactant

To understand this concept, consider the synthesis of water from hydrogen and oxygen:

$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l)$

This balanced equation tells us that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water.

Scenario: Suppose we mix 1 mole of $H_{2}$ and 1 mole of $O_{2}$ .

According to the stoichiometry (2:1 ratio), 1 mole of $H_{2}$ requires $\frac{1}{2}$ mole of $O_{2}$ to react completely.
We have 1 mole of $O_{2}$ available, which is more than the required $\frac{1}{2}$ mole. Therefore, $O_{2}$ is in excess.
For the entire 1 mole of $O_{2}$ to react, it would need 2 moles of $H_{2}$ . We only have 1 mole of $H_{2}$ .
Since the amount of $H_{2}$ is insufficient to react with all the $O_{2}$ , $H_{2}$ is the limiting reactant.
$O_{2}$ is the non-limiting (excess) reactant.

General Procedure to Identify the Limiting Reactant

Calculate the number of moles of each reactant present.
Use the stoichiometry of the balanced equation to calculate the amount of product that could be formed from each reactant independently.
The reactant that produces the least amount of product is the limiting reactant.

Key Point: The reactant with the smallest mass is not necessarily the limiting reactant. It depends on the molar ratio required by the balanced equation and the moles available.

Example 4.11: Synthesis of Calcium Nitride

Calculate the mass of calcium nitride ( $C a_{3} N_{2}$ ) prepared from 54.9 g of Ca and 43.2 g of $N_{2}$ .

The balanced chemical equation is: $3 C a (s) + N_{2} (g) \to C a_{3} N_{2} (s)$

Step 1: Calculate the moles of each reactant.

Molar mass of Ca $= 40.0 g / m o l$
Molar mass of $N_{2} = 2 \times 14.0 = 28.0 g / m o l$

$Moles of Ca = \frac{54.9 g}{40.0 g / m o l} = 1.37 mol$

$Moles of N_{2} = \frac{43.2 g}{28.0 g / m o l} = 1.54 mol$

Step 2: Determine the moles of product each reactant could produce.

From Ca (ratio Ca : $C a_{3} N_{2}$ = 3 : 1): $1.37 mol Ca \times \frac{1 mol C a _{3} N _{2}}{3 mol Ca} = 0.457 mol C a_{3} N_{2}$
From $N_{2}$ (ratio $N_{2}$ : $C a_{3} N_{2}$ = 1 : 1): $1.54 mol N_{2} \times \frac{1 mol C a _{3} N _{2}}{1 mol N _{2}} = 1.54 mol C a_{3} N_{2}$

Step 3: Identify the limiting reactant.

Calcium (Ca) produces the least amount of product (0.457 mol). Therefore, Ca is the limiting reactant and $N_{2}$ is the excess reactant.

Step 4: Calculate the mass of product formed.

Molar mass of $C a_{3} N_{2} = 3 (40.0) + 2 (14.0) = 120.0 + 28.0 = 148.0 g / m o l$

$Mass of C a_{3} N_{2} = 0.457 mol \times 148.0 g / m o l = 67.6 g$

Step 5: Calculate the amount of excess $N_{2}$ remaining.

Moles of $N_{2}$ consumed by 1.37 mol Ca: $1.37 mol Ca \times \frac{1 mol N _{2}}{3 mol Ca} = 0.457 mol N_{2} consumed$

Excess $N_{2}$ remaining: $1.54 - 0.457 = 1.08 mol N_{2} unreacted$

Why Use an Excess Reactant?

In industrial and laboratory processes, chemists deliberately use an excess of one reactant to:

Ensure the more expensive or critical reactant is completely consumed
Maximise the yield of the desired product
Drive the reaction to completion (e.g., excess oxygen in combustion ensures complete burning of fuel)

At Standard Temperature and Pressure (STP) (0°C and 1 atm), one mole of any ideal gas occupies 22.4 dm³ (the molar volume).

This relationship is used to:

Convert between moles and volume of a gas: $n = \frac{V}{22.4 d m ^{3} / m o l}$
Calculate the gram molecular mass of a gas from its density at STP: $M = density \times molar volume = ρ \times 22.4 d m^{3} / m o l$

Example: A gas has a density of $1.25 g / d m^{3}$ at STP. Find its molar mass. $M = 1.25 g / d m^{3} \times 22.4 d m^{3} / m o l = 28.0 g / m o l$ This corresponds to nitrogen gas ( $N_{2}$ ).

Example: Calculate the volume occupied by $4.8 \times 1 0^{24}$ molecules of $C H_{4}$ at STP. $n = \frac{4.8 \times 1 0 ^{24}}{6.022 \times 1 0 ^{23}} \approx 7.97 mol$ $V = 7.97 \times 22.4 \approx 178.5 d m^{3}$

The limiting reactant (or limiting reagent) is the reactant that is completely consumed first in a chemical reaction. The amount of product formed is determined by this reactant.

Identifying the Limiting Reactant

To understand this concept, consider the synthesis of water from hydrogen and oxygen:

$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l)$

This balanced equation tells us that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water.

Scenario: Suppose we mix 1 mole of $H_{2}$ and 1 mole of $O_{2}$ .

According to the stoichiometry (2:1 ratio), 1 mole of $H_{2}$ requires $\frac{1}{2}$ mole of $O_{2}$ to react completely.
We have 1 mole of $O_{2}$ available, which is more than the required $\frac{1}{2}$ mole. Therefore, $O_{2}$ is in excess.
For the entire 1 mole of $O_{2}$ to react, it would need 2 moles of $H_{2}$ . We only have 1 mole of $H_{2}$ .
Since the amount of $H_{2}$ is insufficient to react with all the $O_{2}$ , $H_{2}$ is the limiting reactant.
$O_{2}$ is the non-limiting (excess) reactant.

General Procedure to Identify the Limiting Reactant

Calculate the number of moles of each reactant present.
Use the stoichiometry of the balanced equation to calculate the amount of product that could be formed from each reactant independently.
The reactant that produces the least amount of product is the limiting reactant.

Key Point: The reactant with the smallest mass is not necessarily the limiting reactant. It depends on the molar ratio required by the balanced equation and the moles available.

Example 4.11: Synthesis of Calcium Nitride

Calculate the mass of calcium nitride ( $C a_{3} N_{2}$ ) prepared from 54.9 g of Ca and 43.2 g of $N_{2}$ .

The balanced chemical equation is: $3 C a (s) + N_{2} (g) \to C a_{3} N_{2} (s)$

Step 1: Calculate the moles of each reactant.

Molar mass of Ca $= 40.0 g / m o l$
Molar mass of $N_{2} = 2 \times 14.0 = 28.0 g / m o l$

$Moles of Ca = \frac{54.9 g}{40.0 g / m o l} = 1.37 mol$

$Moles of N_{2} = \frac{43.2 g}{28.0 g / m o l} = 1.54 mol$

Step 2: Determine the moles of product each reactant could produce.

From Ca (ratio Ca : $C a_{3} N_{2}$ = 3 : 1): $1.37 mol Ca \times \frac{1 mol C a _{3} N _{2}}{3 mol Ca} = 0.457 mol C a_{3} N_{2}$
From $N_{2}$ (ratio $N_{2}$ : $C a_{3} N_{2}$ = 1 : 1): $1.54 mol N_{2} \times \frac{1 mol C a _{3} N _{2}}{1 mol N _{2}} = 1.54 mol C a_{3} N_{2}$

Step 3: Identify the limiting reactant.

Calcium (Ca) produces the least amount of product (0.457 mol). Therefore, Ca is the limiting reactant and $N_{2}$ is the excess reactant.

Step 4: Calculate the mass of product formed.

Molar mass of $C a_{3} N_{2} = 3 (40.0) + 2 (14.0) = 120.0 + 28.0 = 148.0 g / m o l$

$Mass of C a_{3} N_{2} = 0.457 mol \times 148.0 g / m o l = 67.6 g$

Step 5: Calculate the amount of excess $N_{2}$ remaining.

Moles of $N_{2}$ consumed by 1.37 mol Ca: $1.37 mol Ca \times \frac{1 mol N _{2}}{3 mol Ca} = 0.457 mol N_{2} consumed$

Excess $N_{2}$ remaining: $1.54 - 0.457 = 1.08 mol N_{2} unreacted$

Why Use an Excess Reactant?

In industrial and laboratory processes, chemists deliberately use an excess of one reactant to:

Ensure the more expensive or critical reactant is completely consumed
Maximise the yield of the desired product
Drive the reaction to completion (e.g., excess oxygen in combustion ensures complete burning of fuel)

At Standard Temperature and Pressure (STP) (0°C and 1 atm), one mole of any ideal gas occupies 22.4 dm³ (the molar volume).

This relationship is used to:

Convert between moles and volume of a gas: $n = \frac{V}{22.4 d m ^{3} / m o l}$
Calculate the gram molecular mass of a gas from its density at STP: $M = density \times molar volume = ρ \times 22.4 d m^{3} / m o l$

Example: A gas has a density of $1.25 g / d m^{3}$ at STP. Find its molar mass. $M = 1.25 g / d m^{3} \times 22.4 d m^{3} / m o l = 28.0 g / m o l$ This corresponds to nitrogen gas ( $N_{2}$ ).