Final answer:
To produce 9.53x10⁻¹ moles of F₂, one would need approximately 74.43 grams of CaF₂ based on stoichiometric calculations using the molar mass of CaF₂.
Step-by-step explanation:
To calculate how many grams of CaF₂ are needed to produce 9.53x10⁻¹ moles of F₂, we must first understand the stoichiometry of the reaction. CaF₂ dissociates into Ca²⁺ and F⁻ ions in solution, and it would take two moles of CaF₂ to produce one mole of F₂ gas because each molecule of CaF₂ contains two fluorine atoms.
Given the molar mass of CaF₂ is 78.08 g/mol, the molecular formula indicates that one mole of CaF₂ would provide two moles of F⁻ ions which could then yield one mole of F₂ gas. Thus:
- For 2 moles of F⁻ from CaF₂, we need 1 mole of CaF₂.
- Therefore, for 9.53x10⁻¹ moles of F₂, we need 9.53x10⁻¹ moles of CaF₂.
- To find the mass required we multiply the moles by the molar mass: 9.53x10⁻¹ moles x 78.08 g/mol gives the mass of CaF₂ required.
Grams of CaF₂ = Moles of CaF₂ x Molar Mass of CaF₂ = 9.53x10⁻¹ mol x 78.08 g/mol = approximately 74.43 grams.