Question

25 g of a Calcium Carbonate sample on decomposition
give 5.5 g CO2. The percentage purity of Calcium
Carbonate in the sample is (atomic weight: Ca = 40)​

Answer 1

Explanation:

CaCO

3

⇌

Δ

CaO+CO

2

Let pure sample of CaCO

3

=x grams

Mass of CaO produced after decomposition =22.4 g

Molar mass of CaCO

3

=100 g/mol

and, Molar mass of CaO=56 g

If 100% is pure, then

100 g CaCO

3

⟶56 g of CaO

Also,y g of CaCO

3

⟶22.4 g of CaO

Dividing these two,

y

100

=

22.4

56

y=40 grams

∴ Percentage of purity =

Totalmassofimpure

Massofpuresample

×100=

50

40

×100=80%

Answer 2

Answer:

50%

Explanation:

CaCO3 on decomposition gives CaO + CO2

--> Molar mass of CaCO3 = 100 g

--> Molar mass of CO2 = 44 g

If 100% is pure then ,

100g CaCO3 ---> 44g of CO2

x g of CaCo3 ---> 5.5g of CO2

So , 100/x = 44/5.5

=> x = 12.5g

% of x = 12.5/25 × 100

= 50%

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