With the knowledge of atomic mass, molecular mass, the mole, the Avogadro’s number and the molar volume, we can make use of the chemical equations in a much better way and can get much useful information from them. Chemical equations have certain limitations as well. They do not tell about the conditions and the rate of reaction. The chemical equations can even be written to describe a chemical change that does not occur. So, when stoichiometric calculations are performed, we have to assume the following conditions. 1. All the reactants are completely converted into the products. 2. No side reaction occurs. Stoichiometry is a branch of chemistry that tells us the quantitative relationship between reactants and products in a balanced chemical equation. While doing calculations, the law of conservation of mass and the law of definite proportions are obeyed. The following type of relationships can be studied with the help of a balanced chemical equation.

1) Mass-mass Relationship

If we are given the mass of one substance, we can calculate the mass of the other substances involved in the chemical reaction.

2) Mass-mole Relationship or Mole-mass Relationship

If we are given the mass of one substance, we can calculate the moles of other substance and vice-versa.

3) Mass-volume Relationship

If we are given the mass of one substance, we can calculate the volume of the other substances and vice-versa. Similarly, mole-mole calculations can also be performed.

Example (11):
Calculate the number of grams of K2SO4 and water produced when 14 g of KOH is reacted with an excess of H2SO4. Also, calculate the number of molecules of water produced.
For doing such calculations, irst of all convert the given mass of KOH into moles and then compare these moles with those of K2SO4 with the help of the balanced chemical equation.
Equation: 2KOH(aq) + H2SO4(aq) K2SO4(aq) + 2H2O(l)
To get the number of moles of K2SO4, compare the moles of KOH with those of K2SO4.

So, 0.125 moles of K2SO4 is being produced from 0.25 moles of KOH Molar mass of K2SO4 = 2 x 39 + 96 = 174 g/mol Mass of K2SO4 produced = No. of moles x molar mass = 0.125 moles x 174 g mol-1 =21.75g

To get the number of moles of H2O, compare the moles of KOH with those of water

So, the number of moles of water produced is 0.25 from 0.25 moles of KOH Mass of water produced = 0.25 moles x 18 g mol-1 = 4.50 g Number of molecules of water = No. of moles x 6.02 x 1023 = 0.25 moles x 6.02 x 1023 molecules per mole = 1.50 x 1023 molecules Answer

Example (2):

Mg metal reacts with HCl to give hydrogen gas. What is the minimum volume of HCl solution (27% by weight) required to produce 12.1g of H2. The density of the HCl solution is 1.14g/cm3. Mg (s) + 2HCl (aq) MgCl2(aq) + H2(g) Solution:
Mass of H2 produced = 12.1 g Molar mass of H2 = 2.016 g mol-1
Mass of H 12.1g Moles of H = Molar mass of H 2.016 mol g = =6.0 moles
To calculate the number of moles of HCl, compare the moles of H2 with those of HCl
So, 12 moles of HCl is being consumed to produce 6 moles of H2. Mass of HCl =Moles of HCl x Molar mass of HCl = 12 moles x 36.5 g mol-1 = 438 grams
2 : HCl 1 : 2 6 : 12 H
2 : HO 2 : 2 1 : 1 0.25 KOH : 0.25
We know that HCl solution is 27% by weight, it means that 27 g of HCl are present in HCl solution = 100 g


Having completely understood the theory of stoichiometry of the chemical reactions, we shift towards the real stoichiometric calculations. Real in the sense that we deal with such calculations very commonly in chemistry. Often, in experimental work, one or more reactants is/are deliberately used in excess quantity. The quantity exceeds the amount required by the reaction’s stoichiometry. This is done, to ensure that all of the other expensive reactants are completely used up in the chemical reaction.

Sometimes, this strategy is employed to make reactions occur faster. For example, we know that a large quantity of oxygen in a chemical reaction makes things burn more rapidly. In this way excess of oxygen is left behind at the end of the reaction and the other reactant is consumed earlier. This reactant which is consumed earlier is called a limiting reactant. In this way, the amount of product that forms is limited by the reactant that is completely used. Once this reactant is consumed, the reaction stops and no additional product is formed. Hence the limiting reactant is a reactant that controls the amount of the product formed in a chemical reaction due to its smaller amount.

The concept of limiting reactant is analogous to the relationship between the number of “kababs” and the “slices” to prepare “sandwiches”. If we have 30 “kababs” and I’ve slices of bread “having 58 slices”, then we can only prepare 29 “sandwiches”. One “kabab” will be extra (excess reactant) and “slices” will be the limiting reactant. It is a practical problem that we can not purchase exactly sixty “slices” for 30 “kababs” to prepare 30 “sandwiches”.
3 : 100 1 g is present in HCl solution = 27 100 438 g are present in HCl solution = x 438 = 1622.2 g 27 Density of HCl solution = 1.14 g/cm Mass of HCl so Volume of HCl =
dilution Density of HCl 1622.2 = 1423 cm Answer 1.14 g g cm−1 = 29

Consider the reaction between hydrogen and oxygen to form water.

2H2(g) + O2(g) 2H2O(l)

When we take 2 moles of hydrogen (4g) and allow it to react with 2 moles of oxygen (64g), then we will get only 2 moles (36 g) of water. Actually, we will get 2 moles (36g) of water because 2 moles (4g) of hydrogen react with 1 mole (32 g) of oxygen according to the balanced equation. Since less hydrogen is present as compared to oxygen, so hydrogen is a limiting reactant. If we would have reacted 4 moles (8g) of hydrogen with 2 moles (64 g) of oxygen, we would have obtained 4 moles (72 g) of water.

Identification of Limiting Reactant

To identify a limiting reactant, the following three steps are performed.

1. Calculate the number of moles from the given amount of reactant.

2. Find out the number of moles of the product with the help of a balanced chemical equation.

3. Identify the reactant which produces the least amount of product as a limiting reactant.

Following the numerical problem will make the idea clear.

Example (13):
NH3 gas can be prepared by heating together two solids NH4Cl and Ca(OH)2. If a mixture containing 100 g of each solid is heated then (a) Calculate the number of grams of NH3 produced. (b) Calculate the excess amount of reagent left unreacted.

2NH4Cl (s) + Ca (OH)2 (s) CaCl2(s) + 2NH3 (g) + 2H2O(l)


(a) Convert the given amounts of both reactants into their number of moles.

Compare the number of moles of NH4Cl with those of NH3 NH4Cl : NH3 2 : 2 1 : 1 1.87 : 1.87 Similarly compare the number of moles of Ca(OH)2 with those of NH3. Ca(OH)2 : NH3 1 : 2 1.35 : 2.70 Since the number of moles of NH3 produced by l00g or 1.87 moles of NH4Cl is less, so NH4Cl is the limiting reactant.

The other reactant, Ca(OH)2 is present in excess. Hence Mass of NH3 produced = 1.87 moles x 17 g mol-1 = 31.79 g Answer

(b) Amount of the reagent present in excess Let us calculate the number of moles of Ca(OH)2 which will completely react with 1.87 moles of NH4Cl with the help of equation. For this purpose, compare NH4Cl and Ca(OH)2
NH4Cl: Ca (OH)2 2: 1
1: 1 2
1.87: 0.935
Mass of NH Cl = 100g Molar mass of NH C1 = 53.5g mol 100 Mass of NH Cl = g = 1.87
53.5g mol Mass of Ca(OH) = 100
Molar mass of Ca(OH) = 74 g mol
Moles of Ca(OH)  g=-1
100 = 1.35 74 g mol =g

Hence the number of moles of Ca(OH)2 which completely react with 1.87 moles of NH4Cl is 0.935 moles. No. of moles of Ca(OH)2 taken =1.35 No. of moles of Ca(OH)2 used = 0.935 No. of moles of Ca(OH)2 left behind = 1.35 – 0.935 = 0.415 Mass of Ca(OH)2 left unreacted (excess) = 0.415×74 = 30.71 g Answer It means that we should have mixed 100 g of NH4Cl with 69.3 g (100 – 30.71) of Ca(OH)2 to get 1.87 moles of NH3.


The amount of the products obtained in a chemical reaction is called the actual yield of that reaction. The amount of the products calculated from the balanced chemical equation represents the theoretical yield. The theoretical yield is the maximum amount of the product that can be produced by a given amount of a reactant, according to the balanced chemical equation. In most chemical reactions the amount of the product obtained is less than the theoretical yield.

There are various reasons for that. A practically inexperienced worker has many shortcomings and cannot get the expected yield. The processes like filtration, separation by distillation, separation by a separating funnel, washing, drying, and crystallization if not properly carried out, decrease the actual yield. Some of the reactants might take part in a competing side reaction and reduce the amount of the desired product. So in most of the reactions, the actual yield is less than the theoretical yield. A chemist is usually interested in the efficiency of a reaction. The efficiency of a reaction is expressed by comparing the actual and theoretical yields in the form of a percentage (%) yield.

Actual yield % yield = x 100 Theoretical yield