• Data

Mass of calcium carbonate (CaCO3) = 20g
Molar mass of calcium carbonate = 100g/mole
Molar volume at S.T.P = 22.4dm3/mole
i. Mass of quicklime produced
ii. Volume of carbon dioxide formed at S.T.P
• The balanced chemical equation for the reaction.

CaCO3(s) CaO(s) + CO2

To convert 20g of CaCO3 into mole

Mole = mass / Molar mass
= 20g / 100g/mole
= 0.2 mole of CaCO3
• i/ Find the mass of quicklime of (CaO)

Make the mole equivalent between CaCO3 and CaO:
1 mole of CaCO3 Ξ 1 mole of CaO
0.2 mole of CaCO3 Ξ ?? of CaO

= (0.2 mole × 1 mole) / 1 mole
= 0.2 mole
The mole of CaO is 0.2 mole

Make conversion of 0.2 mole into mass:
Mass = mole × molar mass of CaO
= 0.2 mole × 56g/mole
= 11.2g

11.2g of quicklime was produced on heating.
• ii/ Find the volume of CO2 produced at S.T.P

from equivalent mole between CaCO3 and CO2:
1 mole of CaCO3 Ξ 1 mole of CO2
0.2 mole of CaCO3 Ξ ?? of CO2

= (0.2 mole × 1 mole) / 1 mole
= 0.2 mole of C02

Convert 0.2 of into volume
Volume = mole × molar volume at S.T.P
= 0.2 mole × 22.4dm3/mole
= 4.48 dm3

4.48dm3 of carbon dioxide gas was produced at S.T.P