Skip to content
Example · Example 1

Q.When 5.6 g5.6\ \text{g} of calcium carbonate (CaCO3\text{CaCO}_3) is heated strongly, it decomposes completely as CaCO3→CaO+CO2\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2, leaving behind 3.14 g3.14\ \text{g} of calcium oxide (CaO\text{CaO}). Calculate the mass of CO2\text{CO}_2 gas released, and show that the law of conservation of mass is obeyed.

West Bengal WbchseTextbookSubjectiveImportance★★★★★est
3% · 1/29 Questions
✓ Free question

By the law of conservation of mass, the total mass of products must equal the total mass of reactants in this closed decomposition.

Mass of CaCO3\text{CaCO}_3 taken =5.6 g= 5.6\ \text{g}

Mass of CaO\text{CaO} formed =3.14 g= 3.14\ \text{g}

Since CaCO3→CaO+CO2\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2, by conservation of mass:

mass of CO2=mass of CaCO3−mass of CaO=5.6−3.14=2.46 g\text{mass of CO}_2 = \text{mass of CaCO}_3 - \text{mass of CaO} = 5.6 - 3.14 = 2.46\ \text{g}

Check using molar masses (M(CaCO3)=100.09M(\text{CaCO}_3) = 100.09, M(CaO)=56.08M(\text{CaO}) = 56.08, M(CO2)=44.01 g mol−1M(\text{CO}_2) = 44.01\ \text{g mol}^{-1}): moles of CaCO3=5.6100.09=0.0560 mol\text{CaCO}_3 = \dfrac{5.6}{100.09} = 0.0560\ \text{mol}, which independently gives mass of CaO=0.0560×56.08=3.14 g\text{CaO} = 0.0560 \times 56.08 = 3.14\ \text{g} and mass of CO2=0.0560×44.01=2.46 g\text{CO}_2 = 0.0560 \times 44.01 = 2.46\ \text{g} — matching exactly.

Since 3.14 g+2.46 g=5.60 g3.14\ \text{g} + 2.46\ \text{g} = 5.60\ \text{g}, the total mass of products equals the total mass of reactant taken.

[!ANSWER] The mass of CO2\text{CO}_2 released is 2.46 g2.46\ \text{g}, and since 3.14 g+2.46 g=5.60 g3.14\ \text{g} + 2.46\ \text{g} = 5.60\ \text{g}, the law of conservation of mass is confirmed.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.