Chemical Equation: Explanation and Examples

Interpretation of Balanced Chemical Equation – Chemical Equation-1 

Key Cocepts

• Chemical equation

• Mass-Mass relationship

• Mass-Volume relationship

Introduction and Explanation: 

‘Stoichiometry’ is derived from the Greek words “stoicheion”(element) and the “metron”(Measure). It is the study of the quantitative aspects of a chemical change. In stoichiometry we try to find the masses, volumes, number of particles, energies, etc., of reactants taking part in a chemical reaction and the products formed. In stoichiometry, we always use a balanced chemical equation (stoichiometric equation) because an unbalanced chemical equation violates the “law of conservation of mass.” 

1 mole of gas=GMW of gas=22.4 L at STP 

1 mole of a substance = GMW 

For homoatomic molecules GMW = Atomicity × GAW 

For heteroatomic molecules, GMW = Sum of GAW of all the constituents. 

A balanced chemical equation gives the following information: 

Example:  

1.  

1. One mole of CH₄ reacts with two moles of O₂ to give one mole of CO₂ and 2 moles of H₂O. 

2. 16 g of CH₄ reacts with 2 × 32 g of O₂ to give 44 g of CO₂ and 2 × 18 g of H₂O. 

3. 22.4 L of CH₄ reacts with 2 × 22.4 L O₂ to give 22.4 L of CO₂ and 2 × 22.4 L of H₂O. 

2.

1. One mole of glucose reacts with six moles of oxygen to give six moles of carbon dioxide and 6 moles of water molecules. 

2. 180 g of glucose reacts with 6×32 g of oxygen to give 6 × 44 g of carbon dioxide and  
   6 × 18 g of water molecule. 

3. 22.4 L of glucose reacts with 6×22.4 L of oxygen to give 6 × 22.4 L of carbon dioxide and 6 × 22.4 L of water molecule. 

Chemical equation: 

A chemical equation is used to represent the chemical reaction by using a formula. A chemical reaction has both the reactants and products. It simply describes the reactants and products that are present in a chemical reaction. A balanced chemical equation provides quantitative data, including the molar ratio at which reactants mix and the molar ratio at which products are created. 

Example:  

When potassium chlorate (KClO₃) is heated it gives potassium chloride(KCl) and oxygen(O₂). 

(Unbalanced chemical equation) 

(Balanced chemical equation) 

A balanced chemical equation contains an equal number of atoms of each element on both sides. 

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. 

2. Calculation based on the mass-mass relationship. 

3. Calculations based on the mass-volume relationship 

4. Calculations based on the volume-volume relationship 

Calculation based on the mass-mass relationship. 

In making some necessary calculations, the following steps are followed. 

1. Write the balanced chemical equation. 

2. Write the theoretical number of reactants and products involved in the reaction. 

3. Calculate the unknown amount of substance using the unitary method. 

Example:  

1. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid according to the reaction. 

How many grams of HCl will react with 5 g of MnO₂? 

Answer: 1 mole of MnO₂ reacts with 4 moles of HCl which means 87g of MnO₂ reacts with                                   146 g of HCl 

5g of MnO₂ reacts with

14687×514687×5

=8.39 of HCl 

2. Calculate the weight of iron which will be corrected into its oxide by the action of 36g of oxygen. (

Answer: Mole ratio of reaction suggest,

Mole of FeMole of H2O=43Mole of FeMole of H2O=43

∴Mole of Fe=43×mole of H2O ∴Mole of Fe=43×mole of H2O 

=43×3632=43×3632

=

32

Weight of Fe =

3. A solid mixture of 5g containing lead nitrate and sodium nitrate was heated below 600^oC until the mass of the residue was constant. If the loss of mass is 30%, find the mass of lead nitrate and sodium nitrate in the mixture. 

(Atomic weight of Pb=207, Na=23, N=14, O=16) 

Answer:

2 × 331=662           2 × 223= 446 

2 × 85                 2 × 69 

Let, weight of Pb(NO₃)₂ in mixture=x  

Weight of NaNO₃ =(5 – x)g 

662g of Pb(NO₃)₂ will give residue=446 

∴∴

xxgm of Pb(NO₃)₂ will give residue =

446662446662

×(x)=0.674x gm×x=0.674x gm

170 g of NaNO₃ give residue=138g 

∴∴

(5 –

xx) f of NaNO₃ will give residue =

138170138170

×(5−x)=0.812(5−x)×5−x=0.8125−x

Actual weight of residue obtained=

(5−5×30100)=3.5g5−5×30100=3.5g

∴0.674x+0.812 (5−x)=3.5∴0.674x+0.812 5−x=3.5

⇒0.138x=0.56⇒0.138x=0.56

x=4.05g=wt of x=4.05g=wt of 

weight of NaNO₃ in the mixture=(5 – 4.05)=0.95 g 

Calculation based on the mass-volume relationship: 

In such calculations, the mass of reactants is given and the volume of the product is required, and vice-versa. 

1 mole of any gas occupies 22.4Litre volume at STP. 

The mass of a gas can be related to volume according to the following ideal gas equation. 

PV=nRT where R = universal gas constant. 

Number of moles(n)=

WeightGram molecular weightWeightGram molecular weight

Values of R = 0.0821L-atm/K-mol, 8.314J/K-mol 

Here P= pressure, V=volume, w=given weight of the substance, and m=Molecular weight of the substance 

Example:  

1. Calculate the volume of NH_3(g) at 27^oC and 1 atm pressure will be obtained by thermal decomposition of 26.75g of NH₄Cl? 

Answer: From the above equation 

1 mole of NH₄Cl= 1 mole of NH₃ 

53.5g of NH₄Cl= 1 mole of NH₃ 

Therefore, 26.75g of NH₄Cl gives

moles of NH₃ 

                         =0.5moles of NH₃ 

PV = nRT 

2. How much marble of 90.5% purity would be required to prepare 10litres of CO₂ at STP when the marble is acted upon by dilute HCl? 

Answer: 

  100 g 22.4 L 

22.4L of CO₂ at STP will be obtained from 100 g of CaCO₃  

∴

10L of CO₂ at STP will be obtained from pure CaCO₃=

10022.4×10=44.64 g 10022.4×10=44.64 g 

∴ 

Impure marble required=

100/90.510090.5

×44.64=49.324 g×44.64=49.324 g

Summary

• ‘Stoichiometry’ is derived from the Greek words “stoicheion”(element) and “metron”(Measure).

• In stoichiometry, we try to find the masses, volumes, number of particles, energies, etc of reactants taking part in a chemical reaction and the products formed.

• For homoatomic molecules GMW = Atomicity x GAW

• For heteroatomic molecules, GMW = Sum of GAW of all the constituents.

• The mass of reactants is given and the volume of the product is required, and vice versa is for the calculation of mass volume relationship

• A chemical equation is used to represent the chemical reaction by using a formula. A chemical reaction has both the reactants and products.