CHEMISTRY | QN 05 | MOLE CONCEPT
20g of carbonate was decomposed on heat to give quicklime and carbon dioxide gas.
Calculate;
i. Mass of quicklime produced.
ii. Volume of carbon dioxide formed at S.T.P
-
Data
Mass of calcium carbonate (CaCO3) = 20g
Molar mass of calcium carbonate = 100g/mole
Molar volume at S.T.P = 22.4dm3/mole
Asked:
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
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Prepared By:
Sir Mahame
Biology & Chemistry teacher,
Mapambano Education Center | Mwenge | DSM.
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