Answer: XeF4= 13.8%, Â XeF6 = 86.2%
Explanation: SOLUTION
fluorine used by the reaction  is 0.0005 mol
Xenon used by the reaction is Â
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 Â
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%