Question

8

Answer 1

Explanation:

Equation -

CaCO₃ → CaO + CO₂

molecular masses of -

CaCO₃ = 40+12+16×3 = 100 g/mol

CaO = 40+16 = 56 g/mol

CO₂ = 12+16×2 = 44 g/mol

Now,

If 100 g of CaCO₃ is decomposed then it produces 56 g of CaO and 44g of CO₂

Then,

If 1 g of CaCO₃ is decomposed then it produces 56/100g of CaO and 44/100g of CO₂

Now,

Here in question, 10g of CaCO₃ is decomposed

It means, It produces

56/100 × 10 = 56/10 = 5.6 g of CaO and

44/100 × 10 = 44/10 = 4.4 g of CO₂

Now,

we get that 10 g of CaCO₃ on burning produces 4.4 g of CO₂

we need to calculate volume of 4.4 g of CO₂

• mole = mass/Gram molecular mass

Then,

mole = 4.4/44 = 0.1

Now,

volume = mole × 22.4 (in Litres)

= 0.1 × 22.4 = 2.24 L

Hence,

2.24 L of CO₂ will be produced when 10g of CaCO₃ decomposed.