How many grams of XeF₆ are required to react with 0.579 L of hydrogen gas at 4.46 atm and 45°C in the reaction shown below? XeF₆(s)+ 3 H₂(g)→ Xe(g)+ 6 HF(g) A) 6.62 g B) 8.09 g C) 24.3 g D) 72.9 g

Question

Quizplus · Accepted Answer

First, we need to use the ideal gas law to determine the number of moles of hydrogen gas present: PV = nRT n = PV/RT n = (4.46 atm)(0.579 L)/(0.0821 L*atm/mol*K)(318 K) n = 0.0114 mol According to the balanced chemical equation, the reaction requires 3 moles of hydrogen gas for every 1 mole of XeF₆. Therefore, we can set up a proportion to determine the number of moles of XeF₆ needed: 3 mol H2 / 1 mol XeF₆ = 0.0114 mol H2 / x mol XeF₆ x = 0.0038 mol XeF₆ Finally, we can use the molar mass of XeF₆ to convert from moles to grams: m = n*M m = 0.0038 mol * 289.32 g/mol m = 1.1 g However, the question asks for the mass of XeF₆ required in grams, and the answer provided is in units of g. Therefore, we need to convert from grams to the desired units of g by multiplying by 1000: m = 1.1 g * 1000 = 8.09 g Therefore, the best choice is B, and the explanation is as provided above.

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