Final answer:
The formula for lead(IV) fluoride is PbF4. To calculate the mass of lead(IV) fluoride formed, you must first determine the moles of reactants and use the balanced chemical equation to find the stoichiometric relationships. The fluoride ion's formula is represented as F-.
Step-by-step explanation:
The formation of lead(IV) fluoride (PbF4) from the reaction of lead(IV) chloride (PbCl4) with fluorine gas (F2) involves a chemical reaction where moles and molar mass are employed to determine the mass of the compound produced. To solve Exercise 6.5.1, you first need to write the balanced chemical equation for the reaction: PbCl4 + 2 F2 → PbF4 + 2 Cl2. Then, you must calculate the molar mass of PbCl4 and use the stoichiometry of the reaction to find out how many moles of PbF4 will be produced. With the molar mass of PbF4, you can then find the mass of lead(IV) fluoride formed.
In response to the question, "Write the formula for each ion", the fluoride ion is represented as F-, as shown in Exercise 4.5.2 and Exercise 3.5.2. Additionally, the relevant exercises and check your learning sections emphasize the concept of neutral ionic compounds, where the sum of the cationic and anionic charges equals zero.