Question

The volume of CO₂ obtained (in litre) at STP by complete decomposition of 100 g CaCO3 sample of 80% purity is

O 1 x 224

0.8 x 11.2

2 x 22.4

0.8 x 22.4​

Answer 1

Explanation:

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Answer 2

Given - CaCO3 - 100 gram with 80% purity

Find - Volume of CO2

Solution - Decomposition reaction of CaCO3 is -

CaCO3 --> CaO + CO2. In the chemical reaction, CaCO3 is calcium carbonate, CaO is calcium oxide and CO2 is carbon dioxide.

Molar mass of CaCO3 is 100 gram/mole.

So, molar concentration of 100 gram CaCO3 = 100/100*80/100

Number of moles = 0.8

As we can see in the reaction, 1 mole of CaCO3 releases 22.4 litre of CO2.

So, 0.8 mole of CaCO3 will release 0.8*22.4

Hence, 0.8 x 22.4 is the correct answer.