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Be sure to answer all parts. Fluorine is so reactive that it forms compounds with materials inert to other treatments. Fluorides of xenon can be formed by direct reaction of the elements at high pressure and temperature. Under conditions that produce only the tetra− and hexafluorides, 1.85 × 10−4 mol of xenon reacted with 5.00 × 10−4 mol of fluorine, and 9.00 × 10−6 mol of xenon was found in excess. What are the mass percents of each xenon fluoride in the product mixture?

Answer :

tutorAnne

Answer: XeF4= 13.8%,  XeF6 = 86.2%

Explanation: SOLUTION

fluorine used by the reaction  is 0.0005 mol

Xenon used by the reaction is  

9.00 x 10^-6 mol- 1.85x10^-4 mol = 0.000176 mol

= 1.76x 10^-4

the two pproducts that can be formed are

Xe + 2F2 -> XeF4 ...........1

Xe + 3F2 -> XeF6 ...........2

let x represent  moles of XeF4  and y represent  mol of XeF6

The sum of mol of Xenon  from the two reactions= 1.76 x 10 ^-4moles

also, the sum of the moles of  Fluorine in nthe two reavctions =0.0005moles

using the coefficients and equating the two reactuons, we have

2x + 3y = 0.0005  

x + y = 0.000176

x= 0.00176-y

2(0.000176 - y) + 3y = 0.0005

0.000352-2y +3y=0.0005

y= 0.0005-0.000352=0.000148 of  XeF6

x + y = 0.000176

x + 0.000148= 0.000176

x= 0.000176-0,000148

x = 0.000028 mol of XeF4

rememeber that mass= no of moles x molar mass

therefore

0.000028 mol of XeF4 x 207.29 g/mol = 0.005804g

0.000148 mol of XeF6 x 245.28 g/mol =0.03630g

% of mass will now be mass/ total mass x 100

% XeF4 =0 .005804 / (.005804 +0 .0363) = 13.8%

% XeF6 =0 .0363g / (0.005804 +0 .0363)g = 86.2%

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