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21 July, 04:03

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)

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  1. 21 July, 07:18
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    8.1433 g of XeF₆ are required.

    Explanation:

    Balanced chemical equation;

    XeF₆ (s) + 3H₂ (g) → Xe (g) + 6HF (g)

    Given dа ta:

    Volume of hydrogen = 0.579 L

    Pressure = 4.46 atm

    Temperature = 45 °C (45+273 = 318 k)

    Solution:

    First of all we will calculate the moles of hydrogen

    PV = nRT

    n = PV / RT

    n = 4.46 atm * 0.579 L / 0.0821 atm. dm³. mol⁻¹. K⁻¹ * 318 K

    n = 2.6 atm. L / 26.12 atm. dm³. mol⁻¹

    n = 0.0995 mol

    Mass of hydrogen:

    Mass = moles * molar mass

    Mass = 0.0995 mol * 2.016 g/mol

    Mass = 0.2006 g

    Now we will compare the moles of hydrogen with XeF₆ from balance chemical equation.

    H₂ : XeF₆

    3 : 1

    0.0995 : 1/3 * 0.0995 = 0.0332 mol

    Now we will calculate the mass of XeF₆.

    Mass = moles * molar mass

    Mass = 0.0332 mol * 245.28 g/mol

    Mass = 8.1433 g
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