LIMITING AND EXCESS REAGENT: HOW TO CALCULATE IT AND EXAMPLES - CHEMISTRY - 2025

The limiting reagent is one that is completely consumed and determines how much mass of products are formed in a chemical reaction; while the reagent in excess is one that does not react completely after having consumed the limiting reagent.

In many reactions, an excess of a reagent is looked for to ensure that all of the reagent of interest reacts. For example, if A reacts with B to produce C, and it is desired that A react completely, an excess of B is added. However, synthesis, and scientific and economic criteria, are what decide whether an excess of A is appropriate. or B.

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The limiting reagent determines the amount of product that can be formed in the chemical reaction. Therefore, if it is known how much of A reacted, it is immediately determined how much of C was formed. Excess reagent never reveals the amounts of product formed.

What if both A and B are consumed in the reaction? Then we speak of an equimolar mixture of A and B. In practice, however, it is not an easy task to ensure that there are equal numbers of moles or equivalents of all reactants; In this case, either of the two, A or B, can be used to calculate the amount of C.

How are the limiting and excess reactants calculated?

There are many ways to identify and calculate the amount of the limiting reagent that can be involved in the reaction. Once calculated, the other reagents are in excess.

A method that allows identifying which is the limiting reagent, based on the comparison of the proportion of reagents with the stoichiometric ratio, is the one described below.

Method 1

A chemical reaction can be outlined in the following way:

aX + bY => cZ

Where X, Y and Z represent the number of moles of each reactant and product. Meanwhile, a, b and c represent their stoichiometric coefficients, resulting from the chemical balance of the reactions.

If the quotient (X / a) and the quotient (Y / b) are obtained, the reactant with the lower quotient is the limiting reactant.

When the indicated ratios are calculated, the relationship between the number of moles present in the reaction (X, Y and Z) and the number of moles involved in the reaction is being established, represented by the stoichiometric coefficients of the reactants (a and b).

Therefore, the lower the quotient indicated for a reagent, the greater the deficit of that reagent to complete the reaction; and therefore, it is the limiting reagent.

Example

SiO ₂ (s) + 3 C (s) => SiC (s) + 2 CO ₂ (g)

3 g of SiO ₂ (silicon oxide) are reacted with 4.5 g of C (carbon).

Moles of SiO ₂

Mass = 3 g

Molecular weight = 60 g / mol

Number of moles of SiO ₂ = 3g / (60g / mol)

0.05 moles

Number of moles of C

Mass = 4.5 g

Atomic weight = 12 g / mol

Number of moles of C = 4.5 g / (12g / mol)

0.375 moles

Quotient between the number of moles of the reactants and their stoichiometric coefficients:

For SiO ₂ = 0.05 mol / 1 mol

Quotient = 0.05

For C = 0.375 moles / 3 moles

Quotient = 0.125

From the comparison of the values ​​of the quotients, it can be concluded that the limiting reactant is SiO ₂.

Method 2

In the previous reaction, the mass produced of SiC is calculated, when 3 g of SiO ₂ is used and when the 4.5 g of C are used

(3 g SiO ₂) x (1 mole SiO ₂ /60 g SiO ₂) x (1 mol SiC / 1 mol SiO ₂) x (40 g SiC / SiC 1 mol) = 2 g of SiC

(4.5 g C) x (3 mol C / 36 g C) x (1 mol SiC / 3 mol C) x (40 g SiC / 1 mol SiC) = 5 g SiC

So, more SiC (silicon carbide) would be produced if the reaction occurred by consuming all the carbon than the amount produced by consuming all of the SiO ₂. In conclusion, SiO ₂ is the limiting reagent, since when all the excess C is consumed, more SiC would be generated.

Examples

-Example 1

0.5 moles of aluminum are reacted with 0.9 moles of Chlorine (Cl ₂) to form aluminum chloride (AlCl ₃): What is the limiting reactant and what is the excess reactant? Calculate the mass of the limiting reagent and the excess reagent

2 Al (s) + 3 Cl ₂ (g) => 2 AlCl ₃ (s)

Method 1

The quotients between the moles of the reactants and the stoichiometric coefficients are:

For aluminum = 0.5 moles / 2 moles

Aluminum quotient = 0.25

For Cl ₂ = 0.9 moles / 3 moles

Cl ₂ quotient = 0.3

Then the limiting reagent is aluminum.

A similar conclusion is reached if the moles of chlorine required to combine with the 0.5 moles of aluminum are determined.

Moles of Cl ₂ = (0.5 mole of the A) x (3 moles of Cl ₂ /2 moles of the A)

0.75 moles of Cl ₂

Then there is an excess of Cl ₂: 0.75 moles are required to react with the aluminum, and 0.9 moles are present. Therefore, there is an excess of 0.15 moles of Cl _2.

It can be concluded that the limiting reagent is aluminum

Calculation of the masses of the reactants

Limiting reagent mass:

Mass of aluminum = 0.5 moles of Al x 27 g / mole

13.5 g.

The atomic mass of Al is 27g / mol.

Mass of excess reagent:

0.15 moles of Cl ₂ remained

Mass of excess Cl ₂ = 0.15 moles of Cl ₂ x 70 g / mol

10.5 g

-Example 2

The following equation represents the reaction between silver nitrate and barium chloride in aqueous solution:

2 AgNO ₃ (aq) + BaCl ₂ (aq) => 2 AgCl (s) + Ba (NO ₃) ₂ (aq)

According to this equation, if a solution containing 62.4g of AgNO ₃ is mixed with a solution containing 53.1 g of BaCl ₂: a) What is the limiting reagent? b) How many of which reactant remain unreacted? c) How many grams of AgCl were formed?

Molecular weights:

-AgNO ₃: 169.9g / mol

-BaCl ₂: 208.9 g / mol

-AgCl: 143.4 g / mol

-Ba (NO ₃) ₂: 261.9 g / mol

Method 1

To apply Method 1, which allows the identification of the limiting reagent, it is necessary to determine the moles of AgNO ₃ and BaCl ₂ present in the reaction.

Moles of AgNO ₃

Molecular weight 169.9 g / mol

Mass = 62.4 g

Number of moles = 62.4 g / (169.9 g / mol)

0.367 moles

Moles of BaCl ₂

Molecular weight = 208.9 g / mol

Mass = 53.1 g

Number of moles = 53.1 g / (208.9 g / mol)

0.254 moles

Determination of the quotients between the number of moles of the reactants and their stoichiometric coefficients.

For AgNO ₃ = 0.367 moles / 2 moles

Quotient = 0.184

For BaCl ₂ = 0.254 moles / 1 mole

Quotient = 0.254

Based on Method 1, the value of the ratios allows to identify AgNO ₃ as the limiting reagent.

Calculation of the mass of the excess reagent

The stoichiometric balance of the reaction indicates that 2 moles of AgNO ₃ react with 1 mole of BaCl _2.

Moles of BaCl ₂ = (0.367 moles of AgNO ₃) x (1 mol BaCl ₂ /2 moles of AgNO ₃)

0.1835 moles of BaCl ₂

And the moles of BaCl ₂ that did not intervene in the reaction, that is, that are in excess, are:

0.254 moles - 0.1835 moles = 0.0705 moles

Mass of BaCl ₂ in excess:

0.0705 mol x 208.9 g / mol = 14.72 g

Summary:

Excess reagent: BaCl ₂

Excess mass: 14.72 g

Calculation of the grams of AgCl produced in the reaction

To calculate the mass of the products, the calculations are made based on the limiting reagent.

g AgCl = (62.4 g AgNO ₃) x (1 mol AgNO ₃ / 169.9 g) x (2 mol AgCl / 2 mol AgNO ₃) x (142.9 g / mol AgCl)

52.48 g

References

1. Whitten, Davis, Peck & Stanley. (2008). Chemistry. (8th ed.). CENGAGE Learning.
2. Flores J. (2002). Chemistry. Santillana Editorial
3. Wikipedia. (2018). Limiting reagent: en.wikipedia.org
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5. Stoichiometry Limiting Reagent Examples. Recovered from: chemteam.info
6. Washington University. (2005). Limiting Reagents. Recovered from: chemistry.wustl.edu