What mass of fluorine gas (F_2) is needed to react completely with 3.6 grams of water (H_2O) according to the following reaction?2 H_2O(l) + 2 F_2(g) → 4 HF(g) + O_2(g)

• A. 9.0 grams
• B. 1.9 grams
• C. 6.8 grams
• D. 3.6 grams

Answer: (C)

Verified step by step guidance

1. Write down the balanced chemical equation: \[2\ H_2O(l) + 2\ F_2(g) \rightarrow 4\ HF(g) + O_2(g)\]
2. Calculate the number of moles of water (H_2O) given its mass. Use the molar mass of water, which is approximately 18.0 g/mol, and apply the formula: \[\text{moles of } H_2O = \frac{\text{mass of } H_2O}{\text{molar mass of } H_2O}\]
3. Use the stoichiometric coefficients from the balanced equation to find the moles of fluorine gas (F_2) needed. According to the equation, 2 moles of H_2O react with 2 moles of F_2, so the mole ratio is 1:1. Calculate moles of F_2 as equal to moles of H_2O.
4. Calculate the mass of fluorine gas required using its molar mass (approximately 38.0 g/mol) and the moles of F_2 found: \[\text{mass of } F_2 = \text{moles of } F_2 \times \text{molar mass of } F_2\]
5. Summarize the process: convert given mass of water to moles, use mole ratio to find moles of fluorine, then convert moles of fluorine to mass. This will give the mass of fluorine gas needed to react completely with 3.6 grams of water.