How many grams of XeF₆ are required to react with 0.579 L of hydrogen gas at 2.46 atm and 45°C in the reaction shown below? XeF₆(s)+ 3 H₂(g)→ Xe(g)+ 6 HF(g) A)3.65 g B)4.46 g C)13.4 g D)40.2 g

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First, use the ideal gas law (PV = nRT) to find the moles of H₂. Given: P = 2.46 atm, V = 0.579 L, R = 0.0821 L·atm/(mol·K), T = 45°C = 318 K. n = PV/RT = (2.46 atm * 0.579 L) / (0.0821 L·atm/(mol·K) * 318 K) ≈ 0.054 mol H₂. From the balanced equation, 1 mole of XeF6 reacts with 3 moles of H₂, so 0.054 mol H₂ would require 0.054 / 3 = 0.018 mol XeF6. The molar mass of XeF6 is approximately 245.29 g/mol (Xe = 131.29, F = 19 * 6 = 114), so the mass required is 0.018 mol * 245.29 g/mol ≈ 4.46 g.

How Many Grams of XeF6 Are Required to React with 0.579

How Many Grams of XeF₆ Are Required to React with 0.579