Question

Lime stone(CaCO3) upon strong heating undergoes thermal decomposition to form lime(CaO) and CO2 gas. The mass of CaO and mass of CO2 formed by the decomposition of 25 g of CaCO3 are (RAM of Ca = 40, C = 12 , O = 16)

Answer 1

Answer:

GMM of CaCO3 = 40 + 12 + 48 = 100 g

GMM of CaO = 40 + 16 = 56 g

GMM of CO2 = 12 + 32 = 44 g

CaCO3 ----> CaO + CO2

1 mole of CaCO3 decomposes to form 1 mole of CaO

so, 100 g of CaCO3 decomposes to form 56 g of CaO

Using unitary method we get,

25 g of CaCO3 decomposes to form 56/4 g of CaO

So, 25 g of CaCO3 decomposes to form 14 g of CaO

Similarly,

1 mole of CaCO3 decomposes to form 1 mole of CO2

so, 100 g of CaCO3 decomposes to form 44 g of CO2

Using unitary method we get,

25 g of CaCO3 decomposes to form 44/4 g of CO2

So, 25 g of CaCO3 decomposes to form 11 g of CO2

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