XeF4= 13.8%, XeF6 = 86.2%
How to mass percentage ?
To calculate the mass percent of an element in a compound, we divide the mass of the element in 1 mole of the compound by the compound's molar mass and multiply the result by 100.
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 products 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 the two reactions =0.0005moles
using the coefficients and equating the two reactions, 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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