## Example:

Calculate the mass of lead chloride formed, by treating an aqueous solution of 6.62g of lead nitrate, with excess of hydrochloric acid.

(Relative atomic mass of: Pb=207, Cl=35.5, H=1, O=16, N=14)

### Solution:

According to the equation,

207 + (14 +16 x 3) x 2 207 + (35.5 x 2)

207 + (14 + 48) x 2 207 + 71

207 + 62 x 2 278

207 + 124

331 g 278 g

331 g of lead nitrate yield 278 g of lead chloride. Hence, mass of lead chloride formed from 6.62 g of lead nitrate is -

Pb(NO3)2 : PbCl2

331 : 278

6.62 : x

Mass of lead chloride formed = 5.56 g

## Example:

A mixture of sodium chloride and anhydrous sodium carbonate has a mass of 5 g. It is dissolved in water, and treated with barium chloride solution. The mass of barium carbonate precipitated is 5.91 g. Calculate the mass of sodium chloride and the percentage of sodium carbonate present in the mixture.

(Relative atomic mass of C=12, O=16, Na=23, Cl=35.5, Ba=137)

### Solution:

According to the equation,

(23 x 2) + {12 + (16 x 3)} 137 + {12 + (16 x 3)}

46 + 12 + 48 137 + 12 + 48

106 g 197 g

Mass of Na2CO3 needed to precipitate 197 g of BaCO3 = 106 g.

Hence, mass of sodium carbonate needed to precipitate 5.91 g of barium carbonate=?

BaCO3 : Na2CO3

197 : 106 g

5.91 g : x

Mass of sodium carbonate present in the mixture = 3.18 g

% of sodium carbonate present in the mixture

Mass of sodium chloride present in the mixture = 5 - 3.18 = 1.82 g

## Example:

Calculate the mass of lead monoxide formed by the complete thermal decomposition of 3.425 g of red lead (Relative atomic mass of O=16, Pb=207).

### Solution:

As per the equation,

(207 x 6) + (16 x 8) (207 x 6) + (16 x 6)

(1242 + 128) 1242 + 96

1370 1338

1370 g of red lead yield 1338 g of lead monoxide.

Mass of lead monoxide formed from 3.425 g of red lead = ?

Pb3O4 : PbO

1370 : 1338

3.425 : x

Mass of lead monoxide formed = 3.345 g

## Example:

Calculate the percentage loss of mass suffered by potassium nitrate, when it is completely decomposed by heat. (Relative atomic mass of K=39, N=4, O=16).

### Solution:

As per the equation,

2 {39 + 14 + (16 x 3)} (16 x 2)

2 (53 + 48) 32

2 (101) 32

202 32

Mass of oxygen liberated when 202 g of KNO3 is heated = 32 g

Percentage loss of mass = 15.84%

## Example:

Calculate the mass of copper nitrate obtained by treating 3.2 g of copper with excess concentrated nitric acid (Relative atomic mass of Cu=64, N=4, O=16).

### Solution:

As per the equation,

64 64 + {14 + (16 x 3)} x 2

64 64 + (62 x 2 )

64 64 + 124

64 188

64 g of copper gives 188 g of copper nitrate.

Hence, mass of copper nitrate formed from 3.2 g of copper = ?

Cu : Cu(NO3)2

64 g : 188 g

3.2 g : x

Mass of copper nitrate formed = 9.4 g