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In this articles we learn class 10 notes science chapter 1. Chemical Reactions and Equations and balancing a chemical equation by hit and trial method.

Notes ⇒ Class 10th ⇒ Science ⇒ 1. Chemical Reactions and Equations

# Notes 1. Chemical Reactions and Equations - Balancing A Chemical Equation | Class 10 Science - Toppers Study

Topper Study classes prepares CBSE Notes on practical base problems and comes out with the best result that helps the students and teachers as well as tutors and so many ecademic coaching classes that they need in practical life. In this articles we learn class 10 notes science chapter 1. Chemical Reactions and Equations and balancing a chemical equation by hit and trial method.

## Notes 1. Chemical Reactions and Equations - Balancing A Chemical Equation | Class 10 Science - Toppers Study

CBSE board students who preparing for class 10 ncert solutions maths and Science solved exercise chapter 1. Chemical Reactions and Equations available and this helps in upcoming exams 2023-2024.

### You can Find Science solution Class 10 Chapter 1. Chemical Reactions and Equations

• All Chapter review quick revision notes for chapter 1. Chemical Reactions and Equations Class 10
• NCERT Solutions And Textual questions Answers Class 10 Science
• Extra NCERT Book questions Answers Class 10 Science
• Importatnt key points with additional Assignment and questions bank solved.

Chapter 1 Science class 10

### Balancing A Chemical Equation class 10 Science Chapter 1. Chemical Reactions and Equations

• Balancing A Chemical Equation By Hit And Trial Method
• Class 10 Science Chapter 1
• Science Class 10 Notes Chapter 1
• Class 10 Science Notes Chapter 1. Chapter 1 Notes Science Class 10
• How To Balance A Chemical Equation
• Balancing Reactants And Products.

## Class 10 notes science chapter 1. Chemical Reactions and Equations

### 1. Balancing a Chemical Equation by hit and trial method.

Here we would learn the method how to balance a chemical reaction.0

The method which we apply e.i hit and trail or inspecting method. Here we take Equation Fe + H2O → Fe3O4 + H2 as example I:

Steps;

(i)  Write each formula in box or imagine that each formula is in box. It is so that       you have to mind no changes should be made inside the box. (ii)  Count and list the numbers of atom of different element that present in                 unbalanced equation. Like this

Elements              Reactant                               Product

Fe                          1                                              3

O                           1                                              4

H                            2                                              2

First observe that what element has the maximum number of atoms. It may be from reactant and from product. Using this criteria we find that Fe3O4  is that compound and element is O that have maximum 4 atoms. To equalise the number of atoms of oxygen, we can put the coefficient 4 with H2O as 4H2O. Then we get the equation as,

Fe + 4H2O → Fe3O4 + H2

(iii)  The next maximum number of atoms having Fe. Which can be equalised

by same process. Putting the coefficient 3 with Fe as 3Fe in reactant side. Now, the equation    will be

3Fe + 4H2O → Fe3O4 + H2

(iv)  Finally we come to equalise Hydrogen atom of both side. We observe that          4H2 in reactant and H2 in product. Total atom of hydrogen in reactant side           is 4 × 2 = 8 and 2 in product. Then, Here we cannot put 2 on LHS and 8 on RHS. (8 ÷ 2 ) =4 , the coefficient will be 4 that equalises hydrogen atom on RHS. So finally equation will be.

3Fe + 4H2O → Fe3O4 + 4H2        (balanced equation)

Example II. We take,

HNO3 + Ca(OH)2 → Ca(NO3)2 + H2O

Steps:

(i)  On observing the above chemical equation we see NO3 has maximum number of atoms/molecules. To equalise it we have on LHS 1 molecule and on RHS 2 molecules. So Here Nitrogen and oxygen have been balanced, then Equation will be;

2HNO3 + Ca(OH)2 → Ca(NO3)2 + H2O

(ii)  Next calcium is already balanced, now we have to balance only hydrogen atoms. Product need 2 as coefficient because (4 ÷ 2) = 2 , then equation will be;

2HNO3 + Ca (OH)2 → Ca(NO3)2 +2H2O

(iii)  On counting and listing the numbers of atom of different element that present in equation above.

Elements              Reactant                   Product

O                           8                                  8

N                           2                                  2

Ca                         1                                  1

H                            4                                  4

Hence we get a balanced equation

2HNO3 + Ca (OH)2 → Ca(NO3)2 +2H2O

I think now you have learnt how to balance a skeletal equation.