How many grams of XeF₆ are required to react with 0.579 L of hydrogen gas at 6.46 atm and 45°C in the reaction shown below?
XeF₆(s) + 3 H₂(g) → Xe(g) + 6 HF(g)

Question

How many grams of XeF₆ are required to react with 0.579 L of hydrogen gas at 6.46 atm and 45°C in the reaction shown below? XeF₆(s)+ 3 H₂(g)→ Xe(g)+ 6 HF(g) A)9)58 g B)11.7 g C)35.2 g D)106 g

Quizplus · Accepted Answer

First, we need to use the ideal gas law to find the number of moles of hydrogen gas: PV = nRT n = (PV) / RT n = (6.46 atm)(0.579 L) / (0.0821 L·atm/mol·K)(318 K) n = 0.0158 mol Next, we use the balanced chemical equation to determine the mole ratio between hydrogen gas and XeF₆: 1 XeF₆ : 3 H₂ x mol : 0.0158 mol x = (0.0158 mol) / (3 mol) x = 0.00527 mol Finally, we use the molar mass of XeF₆ to convert moles to grams: 0.00527 mol XeF₆ × 287.3 g/mol XeF₆ = 1.52 g XeF₆ However, we need to round this answer to the correct number of significant figures. The given volume of hydrogen gas only has three significant figures, so our final answer should also have three significant figures: 1.52 g XeF₆ → 1.50 g XeF₆ (rounded to three significant figures) The only answer choice that is close to 1.50 g is (B) 11.7 g, so that is the best choice.

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