A reaction between calcium fluoride and sulfuric acid produces fluorine gas. If the second sample of calcium fluoride used was 5.00 g, how many grams of fluorine (F_2) are produced, assuming the reaction goes to completion and all calcium fluoride is converted?

• A. 1.26 g
• B. 2.53 g
• C. 5.00 g
• D. 3.79 g

Answer: (B)

Verified step by step guidance

1. Write the balanced chemical equation for the reaction between calcium fluoride (CaF\_2) and sulfuric acid (H\_2SO\_4). The reaction produces fluorine gas (F\_2), calcium sulfate (CaSO\_4), and hydrogen fluoride (HF) or related products depending on the reaction conditions. For this problem, focus on the fluorine-containing products and the stoichiometry involving CaF\_2 and F\_2.
2. Calculate the molar mass of calcium fluoride (CaF\_2). Use the atomic masses: Ca = 40.08 g/mol, F = 18.998 g/mol. The molar mass of CaF\_2 is calculated as: \(M_{CaF_2} = 40.08 + 2 \times 18.998\) g/mol.
3. Determine the number of moles of CaF\_2 in the 5.00 g sample using the formula: \(n = \frac{\text{mass}}{\text{molar mass}} = \frac{5.00 \text{ g}}{M_{CaF_2}}\).
4. Use the stoichiometric ratio from the balanced chemical equation to find the moles of fluorine gas (F\_2) produced from the moles of CaF\_2. For example, if the balanced equation shows that 1 mole of CaF\_2 produces 1 mole of F\_2, then moles of F\_2 = moles of CaF\_2.
5. Calculate the mass of fluorine gas produced by multiplying the moles of F\_2 by its molar mass. The molar mass of F\_2 is \$2 \times 18.998\( g/mol. Use the formula: \)\text{mass of } F_2 = n_{F_2} \times M_{F_2}$.