10.0 g of `CaCO_3` on heating gave 4.4 g of `CO_2` and 5.6g of CaO. The observation is in agreement with the

Options

A. law of constant composition
B. law of multiple proportions
C. law of reciprocal proportion
D. law of conservation of mass

Explanation

To solve the problem, we need to analyze the reaction of calcium carbonate (CaCO₃) when it is heated. The reaction can be summarized as follows:

1. **Identify the Reactants and Products**:
- Reactant: Calcium Carbonate (CaCO₃)
- Products: Calcium Oxide (CaO) and Carbon Dioxide (CO₂)

2. **Write the Balanced Chemical Equation**:
The decomposition of calcium carbonate can be represented by the following equation:
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]

3. **Calculate the Total Mass of Products**:
Given:
- Mass of CaO = 5.6 g
- Mass of CO₂ = 4.4 g
- Total mass of products = Mass of CaO + Mass of CO₂
\[ \text{Total mass of products} = 5.6 \, \text{g} + 4.4 \, \text{g} = 10.0 \, \text{g} \]

4. **Compare the Mass of Reactants and Products**:
- Mass of reactant (CaCO₃) = 10.0 g
- Total mass of products = 10.0 g
- Since the mass of the reactant is equal to the total mass of the products, we can conclude that mass is conserved in this reaction.

5. **Identify the Law that Applies**:
The observation that the mass of reactants equals the mass of products aligns with the **Law of Conservation of Mass**, which states that in a chemical reaction, mass is neither created nor destroyed.

**Final Conclusion**:
The observation is in agreement with the **Law of Conservation of Mass**.

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