## 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 + 71207 + 62 x 2 278

207 + 124331 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(NO_{3})_{2} : PbCl_{2}

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 + 48106 g 197 g

Mass of Na_{2}CO

_{3}needed to precipitate 197 g of BaCO

_{3}= 106 g.

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

BaCO_{3}: Na

_{2}CO

_{3}

197 : 106 g

5.91 g : xMass 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 + 961370 1338

1370 g of red lead yield 1338 g of lead monoxide.Mass of lead monoxide formed from 3.425 g of red lead = ?

Pb_{3}O

_{4}: PbO

1370 : 1338

3.425 : xMass 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) 322 (101) 32

202 32Mass of oxygen liberated when 202 g of KNO_{3} 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 18864 g of copper gives 188 g of copper nitrate.

Hence, mass of copper nitrate formed from 3.2 g of copper = ?Cu : Cu(NO_{3})_{2}

3.2 g : x

Mass of copper nitrate formed = 9.4 g

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