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Chemical Equation: Methods And Examples

Aug 20, 2022
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Key Concepts

• Chemical equation

• Mole-Mole relationship

• Volume-Volume relationship

Introduction

The word Stoichiometry is pronounced as “stoy-key-om-e-tree” which means the calculations of the quantities of reactant and products involved in a chemical reaction. Following are the methods which are used for solving stoichiometric problems.  

  1. Mole method which is for the balanced chemical equation. 

For E.g.,

parallel

Mole of KClO32=MOle of KCl2=Mole of O23Mole of KClO32=MOle of KCl2=Mole of O23

  1. Principle of atom conservation (POAC) method is for unbalanced chemical equation 

For E.g., 

POAC for K:

1×mole of KClO3=1×mole of KCl1×mole of KClO3=1×mole of KCl

POAC for Cl:

parallel

1×mole of KClO3=1×mole of KCl1×mole of KClO3=1×mole of KCl

POAC for O:

3×mole of KClO3=2×mole of O2 3×mole of KClO3=2×mole of O2 

Explanation:

Once we get the balanced chemical equation, we can interpret the chemical equation by the following methods:          

Calculations based on chemical equations are divided into four types: 

  1. Calculation based on the mole-mole relationship. 
  1. Calculation based on the mass-mass relationship. 
  1. Calculations based on the mass-volume relationship 
  1. Calculations based on the volume-volume relationship 
  1. Calculation based on mole-mole relationship: 

In some calculations, if the number of moles of reactants are given then the number of moles of products are calculated, and if the number of moles of products are given, then the number of moles of reactants are calculated. 

Let’s return to the cookie example. Assume I now have plenty of ingredients. The recipe only makes 25 cookies, but I need 100. I’m not sure how to calculate how many cups of chocolate chips I’ll need. I’m not going to count individual chocolate chips. I’m going to do a cup-by-cup calculation. 2 cups of chocolate chips to 25 cookies is my ratio. I want 100 cookies, thus I came up with the following equation: 

(2 cups of chips / 25 cookies) = cups of chips required (100 cookies/1) 

The word ‘cookies’ is crossed out, leaving me with cups of chips as my unit, which is exactly what I’m looking for. I conduct some math (100 * 2) / (25 * 1) and discover that 8 cups of chips are required. Now, we’ll apply the same concept to the chemicals involved in a chemical reaction.  

So, if I want to make 4.5 moles of water, what should I do? How much oxygen do I require in moles? This is an example of a mole-to-mole conversion. You’ll need three pieces of information to complete these calculations: 

  • Find the number of moles in a given substance. 
  • Identity the substances which is to be found. 
  • Mole ratio between given substance and desired substance. Remember, mole ratio comes from the balanced chemical equation! 

Example:  

  1. How many moles of potassium chlorate(KClO3) are required to produce 2.4mole of O2

Solution: 

From the above reaction  

2 moles of KClO3=3 moles of O2 

Since 3 moles of oxygen is formed by 2 moles of KClO3 

Since 2.4 moles of O2 will be formed by

(23×2.4) mole of KClO323×2.4 mole of KClO3

= 1.6 moles of KClO3 

  1. How many moles of carbon dioxide are required when 22.5moles of oxygen is been taken upon the combustion of propane 

Solution: 

From the above reaction of combustion of propane 

5 moles of oxygen = 3 moles of carbon dioxide 

Since it is given that 3 moles of carbon dioxide are formed by 5moles of oxygen 

So, when 22.5moles of oxygen is taken how much carbon dioxide is formed 

=(22.5×53)=4.5 moles of Carbon dioxide=22.5×53=4.5 moles of Carbon dioxide

  1. Calculation based on volume-volume relationship: 

In calculation of volume – volume relationship, the volume of the reactant will be given and you’ll be asked to find the volume that has been produces in the product side. 

Equal volumes of all gases at the same temperature and pressure contain the same number of gas particles, according to Avogadro’s hypothesis. Furthermore, one mole of any gas occupies a volume of 22.4 L at standard temperature and pressure (0°C and 1 atm). Because of these properties, stoichiometry problems involving gases at STP are fairly simple. Consider the formation of nitrogen dioxide by the interaction of nitrogen and oxygen. 

In such calculation, we have Avogadro’s law and Gay-lussac’s law and also one more equation known as the combined gas equation which is given below. 

,where P1, V1, T1 and P2, V2 , T2 are the pressure, volume and absolute (or) Kelvin Temperatures respectively at initial and final conditions. 

Avogadro’s law: 

As this law relates the volume of a gas to the number of molecules at constant temperature and pressure. Avogadro’s law was given by Amadeo Avogadro in the year 1811. This law states that equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules. 

For example: 1 mol of all gases contain 6.023*1023 molecules. At the same time, moles of all gases at 273.15K(0oC) and I bar pressure occupy a volume of the gas is directly proportional to the number of molecules. 

Mathematically we can write as  

V∝NV∝N

( temperature and pressure constant) 

V=Kn, K is the constant of proportionalityV=Kn, K is the constant of proportionality

If m is the mass of a gas having molar mass M, then the number of moles of a gas can be calculated as follows: 

n=mMn=mM

Ideal gas equation: 

An ideal gas is defined as a system in which there are no intermolecular forces and can only exist as a gas. 

The Combination of all the laws that are Boyle’s law, Charles law and are strictly applicable under all conditions of temperatures and pressures. 

From Boyle’s law we get, 

V∝1PV∝1P

                       (At constant n and T) 

From Charles law we get, 

V∝TV∝T

                      (At constant n and P) 

From Avogadro’s law, we get, 

V∝nV∝n

    (At constant T and P) 

Combining the above three equations, we get 

V∝nTPV∝nTP

               or         

V=RnTPV=RnTP          (Where R= ideal gas constant) 

Ideal gas equation is a relation between four variables and it describes the state of any gas. For this reason, it is also called Equation of state. 

Example:  

Calculate the volume of oxygen gas at NTP is necessary for complete combustion of 20 litres of propane measured at 27oC and 760 mm pressure? 

Solution: 

From the above equation 

1Litre of propane requires=5Litres of oxygen 

20Litres of propane requires= 100litres of oxygen 

At given conditions                                                        At NTP conditions 

P1=760 mm of Hg                                                             P2=760 mm of Hg 

V1=100 Litres                                                                    V2=? 

T1=27+273=300 K                                                          T2=273 K 

=91.0 Litres 

                                                      (OR) 

Another method to solve the same question for volume-volume relationship 

Step-I:  

C3H8 at given conditions                                               At NTP 

P1=760mm of Hg                                                            P2=760mm of Hg 

V1=20 Litres                                                                     V2=? 

T1=27+273=300 K                                                         T2=273 K 

=18.2 Litres of C3H8 at NTP 

Step-II: 

1 mole             5moles              3 moles              4moles 

1 mole of C3H8 requires 5moles of O2 

i.e., 1

×22.4 litres of ×22.4 litres of 

C3H8 requires

5×22.4 Litres of 5×22.4 Litres of O2

∴18.2 litres of C3H8 requires=5×22.4×18.222.4

= 91.0litres of O2 of C3H8 requires = 5×22.4×18.2/22.4

Summary

■ The word Stoichiometry is pronounced as “storkey-omp-tree which means the calculations of the quantities of reactant and products involved in a chemical reaction.

■ Ideal gas equation is a relation between four variables and it describes the state of any gas. For this reason, it is also called equation of state.

■ Avogadro’s law was given by Amadeo Avogadro in the year 1811. This law states that equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.

■ Equal volumes of all gases at the same temperature and pressure contain the same number of gas particles,

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