Question

The volume of CO2 evolved at STP
on heating 50g of CaCO3

Answer 1

Answer:

Explanation:

Heating of CaCO₃ is carried out according to following reaction:

CaCO₃ ⇒ CaO + CO₂

Molar mass of CaCO₃ = 100 g

Molar mass of CaO = 56 g

Molar mass of CO₂ = 44 g

According to reaction, 1 mole of CaCO₃ produces 1 mole of CO₂

⇒ 100 grams of CaCO₃ produces 44 grams of CO₂

⇒50 g of CaCO₃ produces (44/100) x 50 = 22 grams

Mass of CO₂ produced = 22/44 = 0.5 moles

Volume of CO₂ = 22.4 x 0.5 moles = 11.2 Liters

Answer 2

Answer:

Heating of CaCO3 is carried out according to following reaction : CaCO3 + CaO + CO2 Molar mass of CaCO3 = 100 g Molar mass of CaO = 56 g Molar mass of CO2 = 44 g According to reaction , 1 mole of CaCO3 produces 1 mole of CO2 100 grams of CaCO3 produces 44 grams of CO2 50 g of CaCO3 produces ( 44/100 ) x 50 = 22 grams Mass of CO2 produced = 22/44 = 0.5 Volume of CO2 = 22.4 x 0.5 moles = 11.2 Liters