Chemistry For Class IX (New Book ) - Chapter No. 1- Fundamentals of Chemistry - Balance Equations And Numericals

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Chapter No.1- Fundamentals of Chemistry
Balance Equations And Numericals
Text Book Exercise

SECTION- D: Numerical

(1) Balance the following equations by inspection method:
(a) NH₃ + O₂ ⟶ NO + H₂O
(b) KNO₃ ⟶ KNO₂ + O₂
(c) Ca + H₂O ⟶ Ca(OH)₂ + H₂
(d) NaHCO₃ ⟶ Na₂CO₃ + H₂O + CO₂
(e) CO + O₂ ⟶ CO₂

Ans: (a) NH₃ + O₂ ⟶ NO + H₂O
Step no.1: Write correct formula of all reactants on left side and product on right side of an equation.

NH₃ + O₂ ⟶ NO + H₂O

Step no.2: Check the number of atoms on each side.

Reactants	Products
N = 1	N = 1
H = 3	H = 2
O = 2	O = 2

Hydrogen is not balance on both side.

Step no.3: (Balance Hydrogen first on both side)
Now multiply the formula (NH₃) with co efficient 4 on reactant side and 6 in front of water (H₂O) on product side to balance the Hydrogen atoms. (if we take coefficient 2, we can not balance oxygen on both side)

4NH₃ + O₂ ⟶ NO + 6H₂O

Reactants	Products
N = 1 x 4 = 4	N = 1
H = 3 x 4 = 12	H =  2 x 6 =12
O = 2	O = 1 x 6 = 6 + 1 = 7

Step no.4: (Balance Nitrogen first than oxygen )
Now again check and balance the equation by placing 4 in front of Nitric oxide (NO) on product side and 5 in front of oxygen on reactant side.

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Reactants	Products
N = 4	N = 1 x 4 =4
H = 12	H = 12
O =  2 x 5 = 10	O =  1 x 4 = 4 + 6 = 10

Now the chemical equation is balanced.

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

(b) KNO₃ ⟶ KNO₂ + O₂
Step no.1: Write correct formula of all reactants on left side and product on right side of an equation.

KNO₃ ⟶ KNO₂ + O₂

Step no.2: Check the number of atoms on each side.

Reactants	Products
K = 1	K = 1
N = 1	N = 1
O = 3	O = 4

Oxygen is not balance.

Step no.3: (Balance Oxygen on reactant side first)
Now multiply the formula of Potassium nitrate (KNO₃) with co-efficient 2 on reactant side

2KNO₃ ⟶ KNO₂ + O₂

Reactants	Products
K = 1 x 2 = 2	K = 1
N = 1 x 2 = 2	N = 1
O = 2 x 3 = 6	O =  4

Step no.4: 
Now multiply the formula of Potassium nitrite (KNO₂) with coefficient 2 on the product side to balance the Potassium, Nitrogen and Oxygen atoms.

2KNO₃ ⟶ 2KNO₂ + O₂

Reactants	Products
K = 1 x 2 = 2	K = 1 x 2 = 2
N = 1 x 2 = 2	N = 1 x 2 = 2
O = 2 x 3 = 6	O =  2 x 2 = 4 + 2 = 6

Now the chemical equation is balanced.

2KNO₃ ⟶ 2KNO₂ + O₂

(c) Ca + H₂O ⟶ Ca(OH)₂ + H₂
Step no.1: Write correct formula of all reactants on left side and product on right side of an equation.

Ca + H₂O ⟶ Ca(OH)₂ + H₂

Step no.2: Check the number of atoms on each side.

Reactants	Products
Ca = 1	Ca = 1
H = 2	H = 4
O = 1	O = 2

Hydrogen and Oxygen are not balanced.

Step no.3: (Balance Oxygen first on reactant side)
Now multiply the formula (H₂O) with co efficient 2 on reactant side.

Ca + 2H₂O ⟶ Ca(OH)₂ + H₂

Reactants	Products
Ca = 1	Ca = 1
H = 2 x 2  = 4	H = 4
O = 1 x 2 = 2	O = 2

Now the chemical equation is balanced.

Ca + 2H₂O ⟶ Ca(OH)₂ + H₂

(d) NaHCO₃ ⟶ Na₂CO₃ + H₂O + CO₂
Step no.1: Write correct formula of all reactants on left side and product on right side of an equation.

NaHCO₃ ⟶ Na₂CO₃ + H₂O + CO₂

Step no.2: Balance the number of atoms on each side.

Reactants	Products
Na = 1	Na = 2
H = 1	H = 2
C = 1	C = 2
O = 3	O = 6

All members are unbalanced.

Step no.3: (Balance Sodium first on reactant side)
Now multiply the formula of Sodium bicarbonate (NaHCO₃) with co efficient 2 on reactant side.

2NaHCO₃ ⟶ Na₂CO₃ + H₂O + CO₂

Reactants	Products
Na = 1 x 2 = 2	Na = 2
H = 1 x 2 = 2	H = 2
C = 1 x 2 = 2	C = 2
O = 2 x 3 = 6	O = 6

Now the chemical equation is balanced.

2NaHCO₃ ⟶ Na₂CO₃ + H₂O + CO₂

(e) CO + O₂ ⟶ CO₂
Step no.1: Write correct formula of all reactants on left side and product on right side of an equation.

CO + O₂ ⟶ CO₂

Step no.2: Check the number of atoms on each side.

Reactants	Products
C = 1	C = 1
O = 3	O = 2

Oxygen is unbalanced.

Step no.3: (Balance Oxygen on product side)
Now multiply the formula of Carbon dioxide (CO₂) with co-efficient 2 on product side.

CO + O₂ ⟶ 2CO₂

Reactants	Products
C = 1	C = 1 x 2 = 2
O = 2	O =  2 x 2 = 4

Step no.4: (Balance Carbon first on the reactant side)
Now again check and balance the equation by placing 2 in front of Carbon monoxide (CO) on reactant side.

2CO + O₂ ⟶ 2CO₂

Reactants	Products
C = 1 x 2 = 2	C = 2
O =  1 x 2 = 2 + 2 = 4	O =  4

Now the chemical equation is balanced.

2CO + O₂ ⟶ 2CO₂

(2) Calculate the formula mass (a.m.u) of the following?
(i) Al₂O₃
(ii) MgCl₂
(iii) NaCl
(iv) KNO₃

i) Al₂O₃
Data:
Formula mass of Al₂O₃ =?
Solution:
Atomic mass of Al = 26.98 a.m.u
Atomic mass of O = 16 a.m.u
Formula unit = Al₂O₃
Formula mass of Al₂O₃ = 2(26.98) + 3(16)
= 53.96 + 69 + 48
= 101.96 a.m.u Ans

ii) MgCl₂
Data:
Formula mass of MgCl₂ =?
Solution:
Atomic mass of Mg = 24.31 a.m.u
Atomic mass of Cl = 35.5 a.m.u
Formula unit = MgCl₂
Formula mass of MgCl₂ = 1(24.31) + 2(35.5)
= 24.31 + 71
= 95.31 a.m.u Ans

iii) NaCl
Data:
Formula mass of NaCl =?
Solution:
Atomic mass of Na = 23 a.m.u
Atomic mass of Cl = 35.5 a.m.u
Formula unit = NaCl
Formula mass of NaCl = 1(23) + 1(35.45)
= 23 + 35.5
= 58.5 a.m.u Ans

iv) KNO₃
Data:
Formula mass of KNO₃ =?
Solution:
Atomic mass of K = 39.10 a.m.u
Atomic mass of N = 14 a.m.u
Atomic mass of O = 16 a.m.u
Formula unit = KNO₃
Formula mass of KNO₃ = 1(39.10) + 1(14) + 3(16)
= 39.10 + 14 + 48
= 101.1 a.m.u Ans

(3) Calculate the molecular mass (a.m.u) of the following?
(i) C₂H₅OH
(ii) H₂O
(iii) NH₃
(iv) CO₂

(i) C₂H₅OH
Data:
Molecular mass (a.m.u) of C₂H₅OH = ?
Solution:
Atomic mass of C = 12 a.m.u
Atomic mass of H = 1 a.m.u
Atomic mass of O = 16 a.m.u
Molecular mass of C₂H₅OH = 2(At. Mass of C) + 6(At. Mass of H) + 1(At. Mass of O)
= 2(12) + 6(1) + 1(16)
= 24 + 6 + 16
= 46 a.m.u Ans

(ii) H₂O
Data:
Molecular mass (a.m.u) of H₂O =?
Solution:
Atomic mass of H = 1 a.m.u
Atomic mass of O = 16 a.m.u
Molecular mass of H₂O = 2(At. Mass of O) + 1(At. Mass of O)
= 2(1) + 1(16)
= 2 + 16
= 18 a.m.u Ans

(iii) NH₃
Data:
Molecular mass (a.m.u) of NH₃ =?
Solution:
Atomic mass of N = 14 a.m.u
Atomic mass of H = 1 a.m.u
Molecular mass of NH₃ = 1(At. Mass of N) + 3(At. Mass of H)
= 1(14) + 3(1)
= 14 + 3
= 17 a.m.u Ans

(iv) CO₂
Data:
Molecular mass (a.m.u) of CO₂ =?
Solution:
Atomic mass of C = 12 a.m.u
Atomic mass of O = 16 a.m.u
Molecular mass of CO₂ = 1(At. Mass of C) + 2(At. Mass of O)
= 1(12) + 2(16)
= 12 + 32
= 44 a.m.u Ans

(4) How many moles are required to prepare 40 gm of H₂SO₄?
Data:
Given mass of H₂SO₄ =40g
Molecular mass of H₂SO₄ = 2(1) + 1(32) + 4(16) = 2 + 32 + 64 = 98 a.m.u
Number of moles =?
Solution:
By using Formula:

(5) Calculate the number of moles and number of molecules present in the following?
(a) 16 g of H₂CO₃
(b) 20 g of C₆H₁₂O₆

(a) 16 g of H₂CO₃
Data:
Number of moles of H₂CO₃ = ?
number of molecules of H₂CO₃ = ?
Given mass of of H₂CO₃ = 16 g
Avogadro’s Number = N_A = 6.02 x 10²³
Solution:
Molecular mass of Carbonoc acid (H₂CO₃) = (2 x 1) + (1 x 12) + (3 x 16)
= 2 + 12 + 48 = 62 a.m.u
Formula:

Number of molecules = Number of moles x N_A
= 0.26 x 6.02 x 10²³
= 1.565 x 10²³ molecules of H₂CO₃ Ans.

(b) 20 g of C₆H₁₂O₆
Data:
Number of moles of C₆H₁₂O₆ = ?
number of molecules of C₆H₁₂O₆ = ?
Given mass of of C₆H₁₂O₆ = 20 g
Avogadro’s Number = N_A = 6.02 x 10²³
Solution:
Molecular mass of glucose ( C₆H₁₂O₆ ) = (6 x 12) + (12 x1) + (6 x 16)
=72 + 12 + 96 = 180 a.m.u
Formula:

Number of molecules = Number of moles x N_A
= 0.111 x 6.02 x 10²³
= 0.668 x 10²³
= 6.68 x 10²² molecules of glucose Ans.

More Numericals
Examples From Text Book

Example 1.1:
If any element have number of protons 11 and number of neutrons 12, find out its atomic number and atomic mass?
Data:
Number of protons = 11
Number of neutrons = 12
Z =?
A =?
Solution:
As we know atomic number Z is number of protons due to this
Atomic number Z = 11
Atomic mass is A = Z + n (Number of neutrons)
A = 11+12
A =23 Ans

Example 1.2:
How many number of protons and neutrons are there in an atom having A= 40 and Z= 20?
Data:
A = 40
Z = 20
Number of protons?
Number of neutrons?
Solution:
As Number of protons is Z = 20
Number of neutrons = A - Z
= 40 - 20
= 20 Ans

Example 1.3:
Calculate the molecular mass of HNO₃ Solution
Data:
Molecular mass of HNO₃ = ?
Solution:
Atomic mass of H = 1 a.m.u
Atomic mass of N = 14 a.m.u
Atomic mass of O = 16 a.m.u
Molecular mass = 1(At. Mass of H) +1(At. Mass of N) +3(At. Mass of O)
= 1 + 14 + 3(16)
= 1 + 14 + 48
= 63 a.m.u Ans

Example 1.4:
Calculate the Formula mass of Al₂(SO₄)₃
Data:
Formula mass of Al₂(SO₄)₃ = ?
Solution:
Atomic mass of Al = 26.98 a.m.u
Atomic mass of S = 32 a.m.u
Atomic mass of O = 16 a.m.u
Formula unit = Al₂(SO₄)₃
Formula mass of Al₂(SO₄)₃ = 2(26.98) + 3(32) +12(16)
= 53.96 + 69 + 192
= 342.14 a.m.u Ans

Example 1.5:
Calculate the number of moles in 40 g of Na.
Data:
Number of moles in 40 g of Na = ?
Solution:
Given mass of Na =40 g
Molecular m ass of Na = 23 a.m.u
Number of moles =?
Formula:

Example 1.6:
What is the mass of 4 moles of CO₂?
Data:
Mass of CO₂ = ?
Solution:
Number of moles of CO₂ = 4 moles
Formula mass of CO₂ = 44 gm
mass of CO₂ = ?
Formula:
Mass of CO₂ = number of moles of CO₂ × formula mass of CO₂
= 4 x 44
= 176 gm

Example 1.7:
Calculate the number of atoms present in 9.2 gm of Calcium (Ca).
Data:
Number of atoms of Ca = ?
Avogadro’s Number = N_A = 6.02 x10²³
Solution:
Atomic mass of Calcium (Ca) = 40
1 g atomic weight of Calcium = 40 gm
40 g of Calcium contains =6.02 x 10²³ atoms of Calcium
Formula:

Example 1.8:
Calculate the number of moles, number of molecules present in 8 g of C₆H₁₂O₆?
Data:
Number of moles of C₆H₁₂O₆ = ?
number of molecules of C₆H₁₂O₆ = ?
Given mass of of C₆H₁₂O₆ = 8 g
Avogadro’s Number = N_A = 6.02 x 10²³
Solution:
Molecular mass of glucose (C₆H₁₂O₆) = (6 x 12) + (12 x 1) + (6 x 16)
= 72 + 12 + 96 = 180 a.m.u
Formula:

Number of molecules = Number of moles x N_A
= 0.04 x 6.02 x 10²³
= 0.240 x 10²³
=2.40 x 10²² molecules of glucose Ans.

Example 1.9:
A coin of silver (Ag) having 8.5 gm weight. Calculate the number of moles of silver in coin?
Data:
Given mass Of Ag = 8.5 gm
Number of moles of Ag = ?
Atomic mass of Ag = 107 a.m.u
Solution:
The mass is converted to number of moles by the following equation:

Example 1.10:
Calculate the number of moles, number of molecules and number of atoms present in 10 gm of H₂SO₄?
Data:
The known mass of H₂SO₄ = 10 gm
Molar mass of H₂SO₄ = 98.0 gm
Number of moles of H₂SO₄ = ?
Number of molecules of H₂SO₄ = ?
Number of atoms of H₂SO₄ = ?
Avogadro’s number = N_A = 6.02 x 10²³
Solution:

Number Of molecules Of H₂SO₄ = Number of moles x Avogadro’s number
= 0.10 x 6.02 x 10²³
= 0.602 x 10²³
= 6.02 x 10²² Ans.

Number of atoms
Number of Atoms Of H₂SO₄ = 2 atoms of H + 1 atoms of S + 4 atoms of O = 7
Number of Hydrogen (H) Atoms in H₂SO₄ = 2/7 x 6.02 x 10²²
= 0.29 x 6.02 x 10²²
= 1.7458 x 10²² atom of H

Number of Sulphur (S) Atoms in H₂SO₄ = 1/7 x 6.02 x 10²²
= 0.143 x 6.02 x 10²²
= 0.86 x 10²²
= 8.6 x 10²¹ atoms of S

Number of Oxygen (O) Atoms in H₂SO₄ = 4 / 7 6.02 x 10²²
= 0.57 x 6.02 x 10²²
= 3.43 x 10²² atoms of O

Example 1.11:
How many liters of carbon dioxide would be produced if 0.450 of a mole of carbon monoxide reacts with excess oxygen at STP.
Data:
Number of moles of carbon monoxide (CO) = x₁ = 0.450 moles
Liter of carbon dioxide = x₂ liters = ?
Solution:
The equation for the reaction is