Part a 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)

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Finnegan

Chemistry

17 March, 21:46

Part a 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)

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Konner Hayes · Answer 1

First, let us find the corresponding amount of moles H₂ assuming ideal gas behavior.

PV = nRT

Solving for n,

n = PV/RT

n = (6.46 atm) (0.579 L) / (0.0821 L-atm/mol-K) (45 + 273 K)

n = 0.143 mol H₂

The stoichiometric calculations is as follows (MW for XeF₆ = 245.28 g/mol)

Mass XeF₆ = (0.143 mol H₂) (1 mol XeF₆/3 mol H₂) (245.28 g/mol) = 11.69 g