Question

'X' gram of calcium carbonate was completely burnt in air. The weight of solid residue formed is \(28 \mathrm{~g}\). What is the value of ' \(\mathrm{X}\) ' (in grams)? (a) 50 (b) 100 (c) 150 (d) 200

Short answer

The value of 'X' is 50 grams.

Step-by-step solution

Understanding the Reaction

When calcium carbonate \(\text{CaCO}_3\) is heated, it decomposes into calcium oxide \(\text{CaO}\) and carbon dioxide \((\text{CO}_2)\). The balanced equation for this reaction is: \[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]

Analyzing Molar Masses

Calculate the molar masses of the chemicals involved in the reaction. - The molar mass of calcium carbonate \(\text{CaCO}_3\) is \(40 + 12 + (3 \times 16) = 100 \ ext{g/mol}\).- The molar mass of calcium oxide \(\text{CaO}\) is \(40 + 16 = 56 \ ext{g/mol}\).

Using Stoichiometry to Relate Reactants and Products

From the balanced chemical reaction:- \(100 \ ext{g}\) of \(\text{CaCO}_3\) gives \(56 \ ext{g}\) of \(\text{CaO}\).- Therefore, the ratio of the mass of calcium carbonate to the mass of calcium oxide is \ \frac{100}{56} \approx 1.786 \.

Calculating Initial Mass of Calcium Carbonate

Given that the mass of the solid residue (calcium oxide) is \28 \ ext{g}\, use the ratio determined in Step 3 to find \(X\), the original mass of calcium carbonate.\[ x = \frac{28 \ ext{g}\ \times 100}{56} = 50 \ ext{g}\ \]

Conclusion

The original mass of calcium carbonate decomposed was \(50 \ ext{g}\). Corresponding to the given options, this suggests the correct value for \(X\) is (a) 50.

Key concepts

Chemical Reactions

In chemistry, chemical reactions are processes where substances, known as reactants, transform into new substances, called products. Reactants undergo chemical changes during this transformation. For example, heating calcium carbonate (\( \text{CaCO}_3 \)) leads to the production of calcium oxide (\( \text{CaO} \)) and carbon dioxide (\( \text{CO}_2 \)) in a decomposition reaction.
This occurs because the bonds between the atoms in the reactants break and new bonds form to create different products. Some key points to remember about chemical reactions:

• Conservation of mass: Mass is neither created nor destroyed in a chemical reaction.
• Reactants and products often differ in their chemical and physical properties.
• Energy changes: Reactions can release or absorb energy, often seen as heat.

When analyzing reactions, writing a balanced chemical equation helps illustrate the changes and lets us predict the quantities involved.

Molar Mass Calculation

Molar mass is a fundamental concept used to convert between the mass of a compound and the number of moles. It's the mass of one mole of a substance, typically expressed in grams per mole (\( \text{g/mol} \)). In stoichiometry, molar mass plays a crucial role in solving chemical equations.For calcium carbonate (\( \text{CaCO}_3 \)), calculating the molar mass involves adding up the atomic masses of its component elements:

• Calcium (Ca): 40 \( \text{g/mol} \)
• Carbon (C): 12 \( \text{g/mol} \)
• Oxygen (O): 3 \times 16 \( \text{g/mol} \)

Therefore, the total molar mass of \( \text{CaCO}_3 \) is 100 \( \text{g/mol} \). Similarly, for calcium oxide (\( \text{CaO} \)), the molar mass is the sum of calcium and oxygen: 56 \( \text{g/mol} \).
Understanding how to calculate these masses aids in determining how much of a reactant or product, in grams, is involved in a reaction.

Balanced Chemical Equations

Balanced chemical equations illustrate how the number of atoms for each element is conserved throughout the reaction process. It’s essential to balance them to satisfy the law of conservation of mass, meaning the mass of the reactants equals the mass of the products.In the decomposition of calcium carbonate:\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \] This equation shows that one mole of calcium carbonate breaks down into one mole of calcium oxide and one mole of carbon dioxide. Here are some steps to ensure a balanced equation:

• Write down the unbalanced equation first.
• Count the number of atoms for each element in both reactants and products.
• Add coefficients to balance the atoms for each element on both sides.

By doing so, predictions and calculations based on the equation will be more accurate, providing a clear view of the stoichiometry involved.