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)

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