Question

Limestone \(\left(\mathrm{CaCO}_{3}\right)\) is decomposed by heating to quicklime \((\mathrm{CaO})\) and carbon dioxide. Calculate how many grams of quicklime can be produced from \(1.0 \mathrm{~kg}\) of limestone.

Short answer

560.2 grams of quicklime can be produced.

Step-by-step solution

Write the Balanced Chemical Equation

The chemical reaction for the decomposition of limestone is: \[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \] This equation shows that 1 mole of \(\text{CaCO}_3\) produces 1 mole of \(\text{CaO}\) and 1 mole of \(\text{CO}_2\).

Calculate Molar Masses

Calculate the molar mass of the compounds involved:- \(\text{CaCO}_3\): \(40.08 + 12.01 + (3 \times 16.00) = 100.09 \, \text{g/mol}\)- \(\text{CaO}\): \(40.08 + 16.00 = 56.08 \, \text{g/mol}\)

Convert Mass of Limestone to Moles

Given 1.0 kg of \(\text{CaCO}_3\), convert to grams: \(\text{mass} = 1000 \text{ g}\).Calculate moles of \(\text{CaCO}_3\): \[ \text{moles of CaCO}_3 = \frac{1000 \text{ g}}{100.09 \, \text{g/mol}} \approx 9.99 \, \text{mol} \]

Use Stoichiometry to Find Moles of CaO Produced

According to the balanced reaction, 1 mole of \(\text{CaCO}_3\) yields 1 mole of \(\text{CaO}\).Therefore, the moles of \(\text{CaO}\) produced are also approximately 9.99 moles.

Convert Moles of CaO to Mass

Convert the moles of \(\text{CaO}\) back into grams using its molar mass:\[ \text{mass of CaO} = 9.99 \, \text{moles} \times 56.08 \, \text{g/mol} \approx 560.2 \, \text{g} \]

Key concepts

Limestone Decomposition

Limestone, also known as calcium carbonate (CaCO_3), is a naturally occurring mineral that undergoes decomposition when heated. This process involves the breaking down of limestone into quicklime (CaO), also recognized as calcium oxide, and carbon dioxide (CO_2). The balanced chemical reaction is:
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]
Here, the solid limestone converts into two products: one solid (CaO) and one gaseous (CO_2). This reaction demonstrates that for every mole of limestone decomposed, one mole of quicklime and one mole of carbon dioxide are produced. This stoichiometric relationship is fundamental and helps us understand the quantitative aspects of the reaction which are crucial for practical applications like producing lime for construction or chemical production.

Molar Mass Calculation

Molar mass is an essential concept in chemistry, allowing us to relate the mass of a substance to the amount in moles. It is calculated by summing the atomic masses of all the atoms in a molecule. This calculation is critical in the context of stoichiometry.
- For limestone ( CaCO_3 ), the molar mass is calculated as: - Calcium ( Ca ): 40.08 g/mol - Carbon ( C ): 12.01 g/mol - Oxygen (3 atoms O ): 3 x 16.00 g/mol - Total = 100.09 g/mol
- For quicklime ( CaO ), the molar mass is: - Calcium ( Ca ): 40.08 g/mol - Oxygen ( O ): 16.00 g/mol - Total = 56.08 g/mol Knowing these molar masses allows us to convert between mass and moles, a crucial step in the stoichiometry of chemical reactions.

Chemical Reactions

Chemical reactions involve the transformation of one or more substances into different substances. These transformations are described by balanced chemical equations, which are foundational to the study of chemistry.
In this context, stoichiometry is the aspect of chemistry that deals with the quantitative relationships between reactants and products in a reaction. When balancing chemical reactions like the decomposition of limestone, we ensure that the number of each type of atom is equal on both sides of the equation.
The decomposition of limestone, as shown in the balanced equation, agrees that one mole of limestone yields one mole of quicklime and one mole of carbon dioxide. This one-to-one relation across reactants and products enables us to predict the amount of product formed from a given amount of reactant, making stoichiometry a powerful tool in both laboratory settings and industrial applications.
By understanding the balanced equation and using molar masses, we can accurately calculate how much quicklime is produced. For example, from 1.0 kg of limestone, we calculate approximately 560.2 grams of quicklime. This practical application demonstrates the importance of chemical reactions and stoichiometry in real-world processes.