When 10.0\text{ g} of pure calcium carbonate (\text{CaCO}_3) is strongly heated, it comple… — JAMB UTME

When $10.0\text{ g}$ of pure calcium carbonate ( $\text{CaCO}_3$ ) is strongly heated, it completely decomposes into solid calcium oxide ( $\text{CaO}$ ) and carbon(IV) oxide gas ( $\text{CO}_2$ ). According to the Law of Conservation of Mass, if $5.6\text{ g}$ of calcium oxide remains in the container, what is the mass of carbon(IV) oxide gas released in grams?

Answer: 4.4 g

Answer

The mass of carbon(IV) oxide gas released is

4.4\text{ g}

According to the Law of Conservation of Mass, the total mass of reactants must equal the total mass of products in a chemical change. For the reaction

\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)

, the initial mass of

10.0\text{ g}

\text{CaCO}_3

must equal the combined mass of

\text{CaO}

(

5.6\text{ g}

) and

\text{CO}_2

. Subtracting

5.6\text{ g}

from

10.0\text{ g}

yields

4.4\text{ g}

for the gas produced.

Step-by-Step Solution

Apply the Law of Conservation of Mass

Total mass of reactants (

10.0\text{ g}

) = Total mass of products (solid residue + gas)

Mass cannot be created or destroyed in a chemical reaction.

Calculate the missing mass of carbon(IV) oxide gas

\text{Mass of CO}_2 = 10.0\text{ g} - 5.6\text{ g} = 4.4\text{ g}

Subtracting the mass of the solid product from the initial mass of reactant gives the mass of the gaseous product evolved.

Key Concept

Law of Conservation of Mass

Estimated Time:45s

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