Given the equilibrium constants (concentration equilibrium constants, units omitted) for the reaction of $\rm{Xe}$ with $\rm{F_2}$ at $500\degree\rm{C}$:

$\rm{Xe+F_2=XeF_2}~~~K_1=29.8$

$\rm{XeF_2+F_2=XeF_4~~~K_2=0.50}$

$\rm{XeF_4}+F_2=XeF_6~~~K_3=3.30\times10^{-3}$

At $500{\degree}\rm{C}$, $\rm{Xe}$ at an initial concentration of $0.100~\rm{mol/L}$ and $\rm{F_2}$ at an initial concentration of $0.200~\rm{mol/L}$ react in a constant-volume container. What is the remaining concentration of $\rm{F_2}$ when the reaction reaches equilibrium? (Retain two significant figures in the result)
已知 $\rm{Xe}$ 与 $ \rm{F_2}$ 在 $500\degree\rm{C}$ 下反应的平衡常数（浓度平衡常数，单位已省略）。

$\rm{Xe+F_2=XeF_2}~~~K_1=29.8$

$\rm{XeF_2+F_2=XeF_4~~~K_2=0.50}$

$\rm{XeF_4}+F_2=XeF_6~~~K_3=3.30\times10^{-3}$

在 $500{\degree}\rm{C}$ 下初浓度 $0.100~\rm{mol/L} ~Xe$ 和 $0.200~\rm{mol/L}$ 的 $\rm{F_2}$ 在恒容容器内反应，则反应达平衡时 $\rm{F_2}$ 的剩余浓度是多少?（结果保留两位有效数字）
All other options are incorrect
0.070
0.080
0.090
0.10
0.11
0.12
0.13
0.14
其他选项均不正确
0.070
0.080
0.090
0.10
0.11
0.12
0.13
0.14
Let the equilibrium concentration of $\rm{Xe}$ be $x$, the concentration of $\rm{F_2}$ be $y$, and the concentrations of $\rm{XeF_2}$, $\rm{XeF_4}$, $\rm{XeF_6}$ be $a$, $b$, $c$ respectively, then:

$x+a+b+c=0.1$

$y+a+2b+3c=0.2$

$a=29.8xy$

<Checkpoint>[$a=29.8xy$](1)</Checkpoint>


$b=0.50ay=29.8\times0.50xy^2$

<Checkpoint>[$b=29.8\times0.50xy^2$](1)</Checkpoint>

$c=3.30\times10^{-3}by=3.30\times10^{-3}\times29.8\times0.50xy^3$

<Checkpoint>[$c=3.30\times10^{-3}\times29.8\times0.50xy^3$](1)</Checkpoint>

Substituting the last three equations into the first two equations, we have:

$x(1+29.8y+14.9y^2+0.0492y^3)=0.1$

$y(1+29.8x+29.8xy+0.1476xy^2)=0.2$

The first equation is transformed into $x=\frac{0.1}{1+29.8y+14.9y^2+0.0492y^3}=\frac{\frac{0.2}{y}-1}{29.8+29.8y+0.1476y^2}$

<Checkpoint>[$x=\frac{\frac{0.2}{y}-1}{29.8+29.8y+0.1476y^2}$](1)</Checkpoint>

Substituting $x$ into the second equation, we obtain $y=0.117$

<Checkpoint>[$y=0.117$](1)</Checkpoint>
设平衡时 $\rm{Xe}$ 浓度 $x$，$\rm{F_2}$ 浓度 $y$，$\rm{XeF_2}$、$\rm{XeF_4}$、$\rm{XeF_6}$ 浓度分别为$a$、$b$、$c$，有：

$x+a+b+c=0.1$

$y+a+2b+3c=0.2$

$a=29.8xy$

<Checkpoint>[$a=29.8xy$](1)</Checkpoint>


$b=0.50ay=29.8\times0.50xy^2$

<Checkpoint>[$b=29.8\times0.50xy^2$](1)</Checkpoint>

$c=3.30\times10^{-3}by=3.30\times10^{-3}\times29.8\times0.50xy^3$

<Checkpoint>[$c=3.30\times10^{-3}\times29.8\times0.50xy^3$](1)</Checkpoint>

将后三式代入前二式，有：

$x(1+29.8y+14.9y^2+0.0492y^3)=0.1$

$y(1+29.8x+29.8xy+0.1476xy^2)=0.2$

第一个式子化为$x=\frac{0.1}{1+29.8y+14.9y^2+0.0492y^3}=\frac{\frac{0.2}{y}-1}{29.8+29.8y+0.1476y^2}$

<Checkpoint>[$x=\frac{\frac{0.2}{y}-1}{29.8+29.8y+0.1476y^2}$](1)</Checkpoint>

$x$ 代入第二个式子，解得 $y=0.117$

<Checkpoint>[$y=0.117$](1)</Checkpoint>