Fluorides of the relatively unreactive noble gas xenon can be formed by direct reaction of the elements at high pressure and temperature. Depending on the temperature and the reactant amounts the product mixture may produce XeF2, XeF4, and XeF6. Under conditions that produce only XeF4 and XeF6, 1.85 E-4 mol of Xe reacted with 5.00 E-4 mol of F2, and 9.00 E-6 mol of Xe was found to be in excess. What are the mass percents of XeF4 and XeF6 in the final product

Mass percent XeF₄: 13.78%

Mass percent XeF₆: 86.22%

Explanation:

Moles of Xe are equal to moles of XeF₄ + Moles XeF₆:

Moles Xe:

1.85x10⁻⁴mol - 9.00x10⁻⁶ mol = 1.76x10⁻⁴ moles Xe reacted:

1.76x10⁻⁴ moles = Moles of XeF₄ + Moles XeF₆ (1)

Moles of F₂ = 1/2 Moles XeF₄ + 1/3 Moles XeF₆

5x10⁻⁴ = 2 Moles XeF₄ + 3 Moles XeF₆ (2)

Replacing (1) in (2):

5x10⁻⁴ = 2 Moles XeF₄ + 3 Moles XeF₆

5x10⁻⁴ = 2 (1.76x10⁻⁴ moles - 1 Mol XeF₆) + 3 Moles XeF₆

5x10⁻⁴ = 3.52x10⁻⁴ - 2 Mol XeF₆ + 3 XeF₆

1.48x10⁻⁴ = Mol XeF₆

And moles XeF₄:

Moles XeF₄ = 1.76x10⁻⁴ moles - 1.48x10⁻⁴ moles = 2.8x10⁻⁵ moles

In grams (Molar mass XeF₄: 207.28g/mol: XeF₆: 245.28g/mol):

Mass XeF₄: 2.8x10⁻⁵ mol * (207.28g / mol) = 5.8x10⁻³g

Mass XeF₆: 1.48x10⁻⁴ mol * (245.28g / mol) = 0.0363g

Mass percent XeF₄:

5.8x10⁻³g / (5.8x10⁻³g + 0.0363g) * 100 = 13.78%

Mass Percent XeF₆:

0.0363g / (5.8x10⁻³g + 0.0363g) * 100 = 86.22%

the answer is 0.8m.

by: nicholas