The problem assumes that the only products of the reaction of $$ \mathrm{Xe} $$ and $$ \mathrm{F}_2 $$ are $$ \mathrm{XeF}_4 $$ and $$ \mathrm{XeF}_6 $$. When 1.85E-4 mol $$ \mathrm{Xe} $$ reacted with 5.00E-4 mol $$ \mathrm{F}_2 $$, there was an excess 9.00E-6 mol $$ \mathrm{Xe} $$. What are the mass % of $$ \mathrm{XeF}_4 $$ and $$ \mathrm{XeF}_6 $$ in the product?
I began by balancing the chemical equation to get
$$ 2\mathrm{Xe}\, +\, 5\mathrm{F}_2\, \rightarrow\, \mathrm{XeF}_4\, +\, \mathrm{XeF}_6$$
At this point I would have thought that 1.76E-4 mol of $$ \mathrm{Xe} $$ (I subtracted the unreacted Xe) had reacted to form 8.8E-5 mol of each product. This will give us .0182 g of $$ \mathrm{XeF}_4 $$ and .0216 g of $$ \mathrm{XeF}_6 $$ with mass percents of 45.7% and 54.3% respectively. The answer in the book said that the products were 14 mass % $$ \mathrm{XeF}_4 $$ and 86.2 mass % $$ \mathrm{XeF}_6 $$. Without any other information I would have expected both products to be formed in the same proportion.