Problem 144 - Chemistry The Science in Context

Question

Preparation of Fluorine Gas HF is prepared by reacting \(\mathrm{CaF}_{2}\) with \(\mathrm{H}_{2} \mathrm{SO}_{4}:\) $$ \mathrm{CaF}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(\ell) \rightarrow 2 \mathrm{HF}(g)+\mathrm{CaSO}_{4}(s) $$ HF can in turn be electrolyzed when dissolved in molten KF to produce fluorine gas: $$ 2 \mathrm{HF}(\ell) \rightarrow \mathrm{F}_{2}(g)+\mathrm{H}_{2}(g) $$ Fluorine is extremely reactive, so it is typically sold as a \(5 \%\) mixture by volume in an inert gas such as helium. How much \(\mathrm{CaF}_{2}\) is required to produce \(500.0 \mathrm{L}\) of \(5 \% \mathrm{F}_{2}\) in helium? Assume the density of \(\mathrm{F}_{2}\) gas is \(1.70 \mathrm{g} / \mathrm{L}\)

Step-by-Step Solution

Answer

Reaction 1: CaF2 + 2H2SO4 -> CaSO4 + 2HF Reaction 2: 2HF + 2Cl2 -> 2HCl + F2 Density of F2: 1.70 g/L Answer: 87.2 g of calcium fluoride (CaF2) is required to produce the 5% fluorine gas mixture.

Step 1: Calculate the amount of F2 in the mixture

The fluorine gas constitutes 5% of the mixture, so we need to calculate the volume of F2 in the 500.0 L mixture. Volume of F2 = (5% of 500.0 L) = 0.05 * 500 = 25 L

Step 2: Convert the volume of F2 to grams

The density of F2 is given as 1.70 g/L. We can use this to convert the volume of F2 to grams. Mass of F2 = Density * Volume = 1.70 g/L * 25 L = 42.5 g

Step 3: Convert the mass of F2 to moles

To work with stoichiometry, we need to convert the mass of F2 to moles. The molar mass of F2 is 2 * Molar mass of Fluorine = 2 * 19 g/mol = 38 g/mol. Moles of F2 = Mass / Molar mass = 42.5 g / 38 g/mol = 1.118 moles

Step 4: Use stoichiometry to find the moles of CaF2 needed

In the first reaction equation, we see that 1 mole of CaF2 produces 2 moles of HF, and the second reaction equation shows that 2 moles of HF produce 1 mole of F2. So, for every mole of CaF2, we get 1 mole of F2. Thus, the moles of CaF2 required is equal to the moles of F2. Moles of CaF2 = 1.118 moles

Step 5: Calculate the amount of CaF2 in grams

Finally, we can convert the moles of CaF2 to grams by using its molar mass. The molar mass of CaF2 is Molar mass of Ca + 2 * Molar mass of F = 40 + 2 * 19 = 78 g/mol. Mass of CaF2 = Moles * Molar mass = 1.118 moles * 78 g/mol = 87.204 g Therefore, 87.2 g of CaF2 is required to produce 500.0 L of 5% F2 in helium.

Key Concepts

Stoichiometry

The principle of stoichiometry is central to predicting the quantities of reactants and products in a chemical reaction. In this exercise, we are focusing on the preparation of fluorine gas from calcium fluoride (\( \text{CaF}_2 \)) and sulfuric acid (\( \text{H}_2\text{SO}_4 \)). Stoichiometry allows us to determine the amount of calcium fluoride needed to produce a specific volume and concentration of fluorine gas.

Firstly, we identify the chemical equations involved. The first equation represents the reaction of \( \text{CaF}_2 \) with \( \text{H}_2\text{SO}_4 \) to produce hydrofluoric acid (HF), which is subsequently electrolyzed to produce fluorine (\( \text{F}_2 \)).

• The equation \( \text{CaF}_2(s) + \text{H}_2\text{SO}_4 (\ell) \rightarrow 2 \text{HF}(g) + \text{CaSO}_4(s) \) indicates that one mole of \( \text{CaF}_2 \) gives two moles of HF.
• The electrolysis equation \( 2\text{HF}(\ell) \rightarrow \text{F}_2(g) + \text{H}_2(g) \) shows that two moles of HF produce one mole of \( \text{F}_2 \).

Thus, it can be concluded that for every mole of \( \text{CaF}_2 \), one mole of fluorine gas is produced. Understanding this ratio allows us to calculate the exact amount of \( \text{CaF}_2 \) that is required based on the moles of \( \text{F}_2 \) desired.

Electrolysis

Electrolysis is crucial in the conversion of hydrofluoric acid (HF) into fluorine gas (\( \text{F}_2 \)). This process uses electrical energy to drive a non-spontaneous chemical reaction. When HF is melted with potassium fluoride (KF), it forms a conductive solution called molten salt, which allows the passage of electric current.

During electrolysis:

• Anode (positive electrode): Oxidation occurs, and HF dissociates to release fluorine gas (\( \text{F}_2 \)).
• Cathode (negative electrode): Reduction occurs, and hydrogen gas is produced.

The balanced reaction in electrolysis is \( 2 \text{HF}(\ell) \rightarrow \text{F}_2(g) + \text{H}_2(g) \). Electrolysis not only allows the isolation of pure fluorine gas but also efficiently separates it from other possible products. It’s a widely used technique in the chemical industry for the generation of reactive gases and pure elements. Understanding the principle of electrolysis helps in comprehending the steps necessary for obtaining fluorine from hydrofluoric acid.

Density Calculation

Density is an essential concept in this exercise as it bridges the gap between grams and liters, which allows us to convert the volume of fluorine gas into mass. The formula is straightforward: \[\text{Mass} = \text{Density} \times \text{Volume}\]Given the density of \( \text{F}_2 \) as 1.70 g/L, we can calculate the mass of fluorine for a given volume.

For instance, if we have 25 liters of \( \text{F}_2 \), then:

• Mass of \( \text{F}_2 \) = 1.70 g/L \( \times \) 25 L = 42.5 g.

Converting the volume to mass enables the use of stoichiometric relationships to find out the required amounts of reactants, such as \( \text{CaF}_2 \) in this instance. Understanding density calculations plays an important role in producing the correct quantity of gases or any substance when given volume constraints.

Chemical Reactions

Chemical reactions are the foundation of this exercise, where we use them to transform reactants into products. The given reactions show both the synthesis of HF from \( \text{CaF}_2 \) and \( \text{H}_2\text{SO}_4 \), and the subsequent electrolysis of HF to yield fluorine. Here, understanding chemical reactions includes recognizing:

• Formation and consumption of substances, like \( \text{CaSO}_4 \) as a by-product in the first reaction.
• The use of stoichiometry to ensure that reactants are present in correct proportions, which avoids wastage and ensures optimal production.
• The flow of a chemical process, starting with raw reactants and leading to the desired end products.

Developing a firm grasp of how chemical reactions take place, and how they are represented through chemical equations, is essential for comprehending how different substances interact and transform. Knowing these details aids in designing and performing chemical syntheses accurately and effectively.