Chapter 3: Problem 80
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
Expert verified
560.2 grams of quicklime can be produced.
Step by step solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]
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.
- 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.
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.
