Chapter 3: Problem 94
Hydrogen fluoride is used in the manufacture of Freons (which destroy ozone in the stratosphere) and in the production of aluminum metal. It is prepared by the reaction $$ \mathrm{CaF}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{CaSO}_{4}+2 \mathrm{HF} $$ In one process, \(6.00 \mathrm{~kg}\) of \(\mathrm{CaF}_{2}\) is treated with an excess of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and yields \(2.86 \mathrm{~kg}\) of \(\mathrm{HF}\). Calculate the percent yield of HF.
Short Answer
Expert verified
The percent yield of HF is approximately 92.86%.
Step by step solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Stoichiometry
Stoichiometry is the area of chemistry that involves calculating the relationships between the reactants and products in chemical reactions. It is rooted in the conservation of mass, meaning that matter cannot be created or destroyed.
In the context of our exercise, stoichiometry helps us determine how much product, in this case hydrogen fluoride (HF), can be produced from a given amount of reactant, calcium fluoride (CaF₂).
The balanced chemical equation, \( \text{CaF}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + 2 \text{HF} \), indicates the proportionate amounts, or moles, of each substance involved in the reaction. Knowing that one mole of \( \text{CaF}_2 \) results in two moles of \( \text{HF} \) aids in finding the theoretical yield when you start with a specific amount of reactant.
In the context of our exercise, stoichiometry helps us determine how much product, in this case hydrogen fluoride (HF), can be produced from a given amount of reactant, calcium fluoride (CaF₂).
The balanced chemical equation, \( \text{CaF}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + 2 \text{HF} \), indicates the proportionate amounts, or moles, of each substance involved in the reaction. Knowing that one mole of \( \text{CaF}_2 \) results in two moles of \( \text{HF} \) aids in finding the theoretical yield when you start with a specific amount of reactant.
Molar Mass
Molar mass is the mass of one mole of any given substance. It's a vital concept for converting grams to moles, which is essential in stoichiometric calculations. Each element’s molar mass is found on the periodic table and represents grams per mole.
For instance, the molar mass of \( \text{CaF}_2 \) is calculated by adding the atomic mass of calcium (\( 40.08\ \text{g/mol} \)) and twice the atomic mass of fluorine (\( 19.00\ \text{g/mol} \)), resulting in \( 78.08\ \text{g/mol} \). Similarly, the molar mass of \( \text{HF} \) is \( 20.008 \text{g/mol} \).
These values allow us to convert the mass of substances into moles and vice versa, a crucial step in determining how much product can theoretically be generated in a chemical reaction.
For instance, the molar mass of \( \text{CaF}_2 \) is calculated by adding the atomic mass of calcium (\( 40.08\ \text{g/mol} \)) and twice the atomic mass of fluorine (\( 19.00\ \text{g/mol} \)), resulting in \( 78.08\ \text{g/mol} \). Similarly, the molar mass of \( \text{HF} \) is \( 20.008 \text{g/mol} \).
These values allow us to convert the mass of substances into moles and vice versa, a crucial step in determining how much product can theoretically be generated in a chemical reaction.
Chemical Reactions
Chemical reactions are processes in which substances (reactants) are transformed into other substances (products). They obey the laws of conservation of mass and energy.
In our reaction, \( \text{CaF}_2 \) reacts with \( \text{H}_2\text{SO}_4 \) to produce \( \text{CaSO}_4 \) and \( \text{HF} \). The reaction shows how matter is conserved, with atoms rearranging rather than disappearing.
Understanding the process, such as what reactants are needed and in what quantities, helps in grasping how much of a product can be expected from a reaction, using stoichiometric calculations to predict outcomes and thereby calculate theoretical yields.
In our reaction, \( \text{CaF}_2 \) reacts with \( \text{H}_2\text{SO}_4 \) to produce \( \text{CaSO}_4 \) and \( \text{HF} \). The reaction shows how matter is conserved, with atoms rearranging rather than disappearing.
Understanding the process, such as what reactants are needed and in what quantities, helps in grasping how much of a product can be expected from a reaction, using stoichiometric calculations to predict outcomes and thereby calculate theoretical yields.
Theoretical Yield
Theoretical yield is the maximum amount of product expected from a chemical reaction, assuming complete conversion of reactants without any loss. It is calculated based on stoichiometric predictions from the balanced equation.
In our example with \( \text{CaF}_2 \) and \( \text{H}_2\text{SO}_4 \), the theoretical yield of \( \text{HF} \) is calculated to be approximately \( 3.08\ \text{kg} \), as deduced from the ideal stoichiometric ratios.
This value acts as a benchmark. Ideally, the actual yield (observed in the lab) should match or come close to the theoretical yield. However, due to various factors like reaction efficiency and practical limitations, the actual yield is often lower, which is why we calculate the percent yield to measure reaction efficiency. By comparing the actual yield to the theoretical yield, we assess the success of the reaction.
In our example with \( \text{CaF}_2 \) and \( \text{H}_2\text{SO}_4 \), the theoretical yield of \( \text{HF} \) is calculated to be approximately \( 3.08\ \text{kg} \), as deduced from the ideal stoichiometric ratios.
This value acts as a benchmark. Ideally, the actual yield (observed in the lab) should match or come close to the theoretical yield. However, due to various factors like reaction efficiency and practical limitations, the actual yield is often lower, which is why we calculate the percent yield to measure reaction efficiency. By comparing the actual yield to the theoretical yield, we assess the success of the reaction.
