Step-by-step 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.