CHEMISTRY, M2 EQ-Bank 2

In an experiment, calcium carbonate \(\ce{(CaCO3)}\) is heated strongly to produce calcium oxide \(\ce{(CaO)}\) and carbon dioxide according to the reaction below:

\(\ce{CaCo3(s) -> CaO(s) + CO2(g)}\)

A student starts with 50.0 g of calcium carbonate. After heating, they collect 28.0 g of calcium oxide.

Using the law of conservation of mass, calculate the mass of carbon dioxide gas produced in this reaction. (2 marks)

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Explain how the law of conservation of mass applies to this reaction. (2 marks)

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Show Answers Only

a. \(22.0\ \text{g}\)

b. The law of conservation of mass states that the mass of reactants in a chemical reaction must equal the mass of the products.

→ Here, the 50.0 g of calcium carbonate decomposes into 28.0 g of calcium oxide and 22.0 g of carbon dioxide gas.

→ The total mass of products (28.0 g + 22.0 g) equals the initial mass of reactants (50.0 g), confirming that mass is conserved in this reaction.

Show Worked Solution

a. According to the law of conservation of mass, the total mass of reactants must equal the total mass of products.

Since mass must be conserved:

\(m\ce{(CO2)}\)	\(=m\ce{(CaCO3)}-m\ce{(CaO)}\)
	\(= 50.0-28.0\)
	\(=22.0\ \text{g}\)

b. The law of conservation of mass states that the mass of reactants in a chemical reaction must equal the mass of the products.

→ Here, the 50.0 g of calcium carbonate decomposes into 28.0 g of calcium oxide and 22.0 g of carbon dioxide gas.

→ The total mass of products (28.0 g + 22.0 g) equals the initial mass of reactants (50.0 g), confirming that mass is conserved in this reaction.