How to determine limiting reagent or reactant in a reaction

it s very important to learn how to determine the limiting reagent in industries and laboratories in order to deduct the amount of theoretical yields podced.t.

To deduce the limiting reagent in a chemical reaction, you need to remember that the limiting reagent is the reactant that finishes first in a reaction and determines the amount of product formed in a reaction.

The limiting reactant (limiting reagent) is the reactant among the reactants that is used up first and restricts the yield of the product.

The theoretical yield of a product in a chemical reaction is dependent on the limiting reagent or reactant of a reaction and not the excess reactant.

It takes two or more reactants to produce a product in a chemical reaction, and if the reagents are not inaccurate stoichiometric amounts then one reactant will be in excess while the other reagent(s) will be in excess.

Common Questions on limiting and excess reactant

What is the difference between limiting reagents and limiting reactants?

There is no difference between limiting reagents and limiting reactants. Actually, the limiting reagent is another name for the limiting reactant. You can see that I used the two terms interchangeably just to emphasize that the limiting reagent is the same as the limiting reactant.

What is limiting and excess reagent?

Please note that the limiting reagent is the reactant that is completely used up in a chemical reaction and determines the amount of product while the excess reactant is the reactant that remained after the reaction.

Additionally, the limiting reagent produces the least amount of product hen compared to the excess reactant..

How does limiting reagent affect theoretical yield?

The amount of products that can be calculated from the stoichiometry of chemical equations which is otherwise called the theoretical yield varies directly with the limiting reagent.

So if the limiting reagent is available, the theoretical yield will still be increasing but ff the limiting reagent finishes the theoretical yield finishes.

What happens when the limiting reactant n a chemical reactant is completely used up?

When the limiting reactant is completely used up, the reaction will stop because there is no reaction happening again.

What happens when the limiting reactant runs out is that the amount of product formed will remand constant.

Is it possible to have no limiting reagent?

Yes, there is no limiting reagent when the quintiles are in pope stoichiometric amounts. T means that everything is prepared according to the mole ratio.

How to find limiting and excess reactants?

There are three basic ways to solve the limiting and excess reactant problems; limiting reactant with grams, limiting reactant with moles, and limiting reactants with volumes (for gaseous reactants and products)

How do I deduce the limiting reagent in a chemical reaction?

How do you find the limiting reagent in a chemical reaction? A limiting reactant or reagent is the reactant that is finished during the reaction. The simplest way to analyze this is to check the amount that gave zero at the end of the reaction.

Why is the limiting reagent impotent?

The limiting reagent is extremely important because it helps to deduce the stoichiometric amount or quantity of product that can be produced by inspecting and solving a chemical exaction.

How to find how much limiting reagent is left over?

This is just like asking for the amount of meaning excess reactant that has not been used. The best way to do this is to subtract the limiting reagent amount or reacted amount from the excess reagent and then you will get the amount of reagent left over.

Some examples of limiting reagent problems and answers

How to find limiting reactants with grams

So I want to use a few examples to show how to calculate the limiting reactant with gram that the mass of the reactants.

Example 1

Consider the reaction Zn + 2HCl ==ZnCl₂ + H₂. If 80g of Zinc reacted with 35g of hydrochloric acid, what is the limiting and excess reactant?

Solution

. Zn + 2HCl ==ZnCl₂ + H₂

Recall, we mentioned that the limiting reagent is the reagent with the least number of moles or the reactant that will yield the least amount of theoretical yield.

From the periodic table, pick the atomic mass of the elements involved.

Zinc is 65, Hydrogen is 1, and Chlorine is 35.5

Molar /atomic mass of Zn =65    Molar mass of HCl = 1 +35.5 =36.5

From the balanced equation

Fest calculate the number of moles using n =m/M ; number of moles = mass/molar mass

1 mole of Zn  : 2 moles of HCl

n=80/65 = 1.23 noles       n= .35/36.5 =0.959 mole

Then divide by the coefficients of the reactants   

1.23/1 =1.23 moles         0.959/2=0.48 mole

So the limiting reagent is hydrochloric acid because it produces the least number of moles.

Alternatively, you can check the amount of products that the two reactants can produce by comparing the number of moles of the reactants to a specific product. The limiting reagent will produce the smallest theoretical yield.

Comparing Zinc(reactant) and Zinc Chloride (product)

1 mole of Zn =======1 mole of ZnCl₂

1.23 moles of Zn ======?

   Mole of ZnCl₂ = 1.23 moles

Comparing Hydrochloric acid(reactant) and Zinc Chloride(product)

2 moles of Hydrochloric acid (HCl)===== 1 mole of Zinc Chloride (ZnCl₂)

0.959 moles of hydrochloric acid =========?

Moles of Zinc Chloride = 0.48 moles of Zinc Chloride

So you can still verify that hydrochloric acid while the excess reactant is Zinc.

Example 2

If 36g of Magnesium reacted with 78g of Zinc chloride according the equation below:

Mg + ZnCl₂ === MgCl₂ +Zn

Solution

Mg + ZnCl₂ === MgCl₂ +Zn

From the periodic table, pick the atomic mass of the elements involved.

Magnesium is 24, Zinc is 65, Hydrogen is 1, and Chlorine is 35.5

Molar /atomic mass of Mg =24    Molar mass of ZnCl₂ = 65 +35.5 x 2 =136

From a balanced chemical equation

1 mole of Mg: 1 mole of ZnCl₂

n=36/24 =1.5 mole       n= .78/136=0.574 mole

Then divide by the coefficients of the reactants   

1.5/1 =1.5 moles         0.574/1=0.574 mole

So the limiting reagent is the Zinc Chloride because it produces the least number of moles.

Alternatively, we can still check the amount of products that the two reactants can produce by comparing the number of moles of the reactants to a specific product. The limiting reagent will produce the smallest theoretical yield.

Okay, let’s try it again

Comparing Magnesium (reactant) and Zinc Chloride (product)

1 mole of Mg =======1 mole of ZnCl₂

1.5 moles of Mg ======?

   Mole of ZnCl₂ = 1.5 moles

Comparing Zinc Chloride (reactant) and Magnesium Chloride (product)

2 moles of Hydrochloric acid (HCl)===== 1 mole of Zinc Chloride (ZnCl₂)

0.574 moles of hydrochloric acid =========?

Moles of Zinc Chloride = 0.287 moles of Magnesium Chloride

So you can still verify that Zinc Chloride is the limiting reagent while the excess reactant is Magnesium.

How to find limiting reagent with moles

Example 1

If 2 moles of propane reacts with 6 moles of oxygen, determine the limiting reagent.

C₃H₈ + 5O₂ === 3CO₂ + 4H₂O

Solution

C₃H₈ + 5O₂ === 3CO₂ + 4H₂O

 I have two reactants I am considering

1 mole  :  5 moles

 Divide the mole by the coefficients n the balanced chemical equations.

2/1 = 2 moles   6/5 =1.2 moles

The reactant with the smallest number of moles is the limiting reactant because it is the reactant that runs out.

Alternatively

You can compare the reactant to the theoretical yield, the limiting reagent will produce the smallest yield of product which is the theoretical yield.

. C₃H₈ + 5O₂ === 3CO₂ + 4H₂O

Using the number of moles of propane by comparing propane to carbon IV oxide

1 mole of C₃H₈ yields 3 moles of carbon iv oxide

2 moles of C₃H₈ yields =?

Therefore 2 x 3/1 = 6 moles of carbon IV oxide.

Let’s check the second reactant oxygen

If 5 moles of oxygen produces 3 moles of carbon IV oxide

6 moles of oxygen produces =?

Therefore 6 x 5/3 = 3.6 moles of carbon IV oxide

So the limiting reagent or reactant is the reagent that runs out or produces the least theoretical yield.

Example 2

If 3 moles of Carbon monoxide reacts with 3 moles of oxygen to produce carbon IV oxide according to the equation below 2CO + O₂ ===2CO₂. Determine the limiting reagent.

Solution

2CO + O₂ === 2CO₂

3 moles of CO: 3 moles of O₂

Remember the one with the least number of moles is the limiting reagent.

3/2 = 1.5 moles of CO   3/1 = 3 moles of O₂

So the limiting reagent here is the Carbon monoxide gas.

Off course, you can stall verify using the least amount of product formed which is the theoretical yield of the product. The reactant that produced the least theoretical yield is the limiting reagent.

Alternatively

2CO + O₂ === 2CO₂

You can compare the reactant to the theoretical yield, the limiting reagent will produce the smallest yield of product which is the theoretical yield.

Using the number of reactants by comparing the reactants to the products:  comparing Carbon monoxide to carbon IV oxide

2 moles of CO yields==== 2 moles of carbon IV oxide

1.5 moles of CO yields =====?

Moles of carbon IV oxide gas = 1.5 x 2 /2 = 1.5 moles of Carbon IV oxide

Let’s check the second reactant oxygen

If 1 mole of oxygen produces== 2 moles of carbon IV oxide

3 moles of oxygen produces =?

Therefore 3 x3 = 9 moles of carbon IV oxide

So the limiting reagent or reactant is the reagent runs out or produces the least theoretical yield which n this case is still the carbon IV oxide

How to find limiting reagent without mass

I am going to show us a way to find the limiting reagent without mass; actually, sometimes e can be asked to deduce the limiting reagent by other quantities except for mass and moles.

The concentration of reactants is the best option usually considered in this concept and we will look at the examples on concentration.

The common quantity used is the concentration of the reactants but this is still very easy and I believe we can navigate our way through it with ease.

Example 1

If 3.5g of Magnesium reacted with 0.04 mol/dm³ of 240 cm³ Titanium IV Chloride, Calculate the limiting reagent.

2Mg  + TiCl₄ ===2MgCl₂ + Ti

Solution

Yes, I used a mass for the solid reactant but I need to show how we can solve limiting reagents without mass   

Calculate the number of moles from the balanced equation

Number of moles of Mg = mass of Mg/Molar mass

Number of moles = 3.5/24 =0.146 mole

Number of moles of TiCl₄= concentration x volume in dm³

Divide by the coefficient mole of magnesium = 0.146/2 =0.073 mole

Number of moles = conc. x volume

                             = 0.04 x (240/1000)=0.0096 mol

Divide by the coefficient mole of titanium IV chloride =0.0096/1 =0.0096 mole

2Mg  + TiCl₄ ===2MgCl₂ + Ti

Conclusively, the limiting reagent is the one with the least number of moles which is TiCl₄ for this equation.

Example 2

What is the limiting reagent between the reaction of 0.87 mol/dm³ of 300cm³ HCl and 0.65 mol/dm³ of 280 cm³ NaOH?

HCl + NaOH ===NaCl + H₂O

Solution

Calculate the number of moles HCl = concentration x volume in dm³

 Calculate the number of moles HCl=0.87 x (300/1000)

                                                          =0.261 mole of HCl

Calculate the number of moles NaOH= concentration x volume in dm³

 Calculate the number of moles NaOH=0.65 x (280/1000)

                                                          =0.182 moles of HCl

So the limiting reagent in this neutralization reaction is the hydrochloric acid because it gave the least number of moles.

How to find limiting reagent with volume

We can also find limiting reactant with volume if the reactants are gaseous because volume and pressure are better used to analyze gases.

Example 1

If 20cm³ of nitrogen reacted with 30cm³ of hydrogen reacted to form ammonia, determine the limiting and excess reactant.

N₂ + 3H₂ ===2NH₃

Solution

N₂ + 3H₂ ===2NH₃

In this case, e can use the volume of gas at room temperature or the volume of a gas at standard temperature and pressure

The volume of a gas at standard temperature and pressure is 22.4 dm³

The volume of a gas at room temperature and pressure is 24 dm³

Let me use the mola volume – 22.4 dm³ =22400cm³

Calculate the number of moles N₂= given volume /molar volume

 Calculate the number of moles N₂= 20/22400

                                                          =0.00089 mole

Divide by the coefficients of the nitrogen reactant = 0.00089 mole/1 = 0.00089 mole

Calculate the number of moles H₂= given volume /molar volume

Calculate the number of moles H₂= 30/22400

                                                              =0.00134 mole

Divide by the coefficients of the hydrogen reactant = 0.00134mole/3 = 0.00045 mole

So the limiting reagent in this reaction is the hydrogen because it gave the least number of moles.

How do I find how much limiting reagent is left over? This is the same as asking or saying how do I find the excess reactant leftover; it is very simple, just subtract the limiting reagent from the excess reactant

A proper understanding of some examples of limiting reactant problems and answers will serve as a practical guide to solving questions solving analysis and calculations on limiting reagents and theoretical yield.