$\mathbf{X}$ is a basic carbonate of metal $\mathbf{M}$, possibly containing water of crystallization. Two samples of equal mass were taken and treated separately: one was thoroughly treated in hydrogen, collecting $2.16$ $\mathrm{g}$ of water, and the remaining residue was the element of metal $\mathbf{M}$, weighing $10.36$ $\mathrm{g}$; the other was treated in oxygen, and after complete reaction, $0.448$ $\mathrm{L}$ of carbon dioxide was obtained ($101.325$ $\mathrm{kPa}$, $0 ^ { \circ } \mathrm { C } )$, and the remaining residue was the highest oxide of $\mathbf{M}$. Then the metal $\mathbf{M}$ and the carbon:oxygen ratio in $\mathbf{X}$ are respectively:
$\mathbf{X}$是金属$\mathbf{M}$的碱式碳酸盐，可能含结晶水。称取质量相等的样品两份，分别进行处理：一份在氢气中充分处理，收集到 $2.16$ $\mathrm{g}$ 水，剩余残渣为金属$\mathbf{M}$的单质，重$10.36$ $\mathrm {g}$ ；另一份则在氧气处理，完全反应后得$0.448$ $\mathrm{L}$二氧化碳（$101.325$ $\mathrm{kPa}$， $0 ^ { \circ } \mathrm { C } )$ ，剩余残渣为$\mathbf{M}$的最高价氧化物。则金属$\mathbf{M}$以及$\mathbf{X}$中的碳:氧比例分别为:
All other options are incorrect.
Cu,1:5
Cu,1:8
Cu,4:15
Mn,1:5
Mn,1:8
Mn,4:15
Zn,1:5
Zn,1:8
Zn,4:15
Pb,1:5
Pb,1:8
Pb,4:15
U,1:5
U,1:8
U,4:15
La,1:5
La,1:8
La,4:15
其他选项均不正确
Cu,1:5
Cu,1:8
Cu,4:15
Mn,1:5
Mn,1:8
Mn,4:15
Zn,1:5
Zn,1:8
Zn,4:15
Pb,1:5
Pb,1:8
Pb,4:15
U,1:5
U,1:8
U,4:15
La,1:5
La,1:8
La,4:15
Assuming the chemical formula of $\mathbf{X}$ is $\mathbf{M}\mathrm{(OH)_x(CO_3)_y\cdot zH_2O}$,

<Checkpoint>[Set up unknowns to represent the chemical formula of $\mathbf{X}$ as $\mathbf{M}\mathrm{(OH)_x(CO_3)_y\cdot zH_2O}$](1)</Checkpoint>

then the equations for the two reactions are respectively:

$\mathbf{M}\mathrm{(OH)_x(CO_3)_y\cdot zH_2O+(\frac{x}{2}+y)H_2\to yCO_2+(x+y+z)H_2O}+\mathbf{M}$

and

$\mathbf{M}\mathrm{(OH)_x(CO_3)_y\cdot zH_2O+(\frac{2a-x-2y}{4})O_2\to (\frac{x}{2}+z)H_2O+yCO_2+}\mathbf{M}O_a$

Thus, we obtain the equation:

$\frac{x+y+z}{y}=\frac{2.16/18.016}{0.448/22.4}$

Simplifying, we get $x+z=5y$

<Checkpoint>[Deduce the relationship between unknowns $x+z=5y$](1)</Checkpoint>

From the mass of the metal, we can list the second equation:

$\frac{10.36}{\mathrm{M}}=\frac{2.16}{18.016(x+y+z)}=\frac{2.16}{18.016\cdot (6y)}=\frac{1}{50.04y}$

Try to solve the equation. When $y=2/5$, there is a reasonable solution $\mathrm{Pb}$.

<Checkpoint>[Obtain $\mathbf{M}$ as $\mathrm{Pb}$ by solving the indeterminate equation](1)</Checkpoint>

At this time, $x+z=5y=2$

The common valence states of $\mathrm{Pb}$ are +2 and +4. Here, it is obviously +2, so from the valence, we can calculate $x=\frac{6}{5}$, thus $z=\frac{4}{5}$

Therefore, $\mathbf{X}$ should be $\mathrm{Pb(OH)_{\frac{6}{5}}(CO_3)_{\frac{2}{5}}\cdot \frac{4}{5}H_2O}$.

<Checkpoint>[The chemical formula of $\mathbf{X}$ is $\mathrm{Pb(OH)_{\frac{6}{5}}(CO_3)_{\frac{2}{5}}\cdot \frac{4}{5}H_2O}$](1)</Checkpoint>

Therefore, choose option L
假定$\mathbf{X}$的化学式为$\mathbf{M}\mathrm{(OH)_x(CO_3)_y\cdot zH_2O}$，

<Checkpoint>[设出未知数将$\mathbf{X}$的化学式表示为$\mathbf{M}\mathrm{(OH)_x(CO_3)_y\cdot zH_2O}$](1)</Checkpoint>

则两个反应的方程式分别为：

$\mathbf{M}\mathrm{(OH)_x(CO_3)_y\cdot zH_2O+(\frac{x}{2}+y)H_2\to yCO_2+(x+y+z)H_2O}+\mathbf{M}$

和

$\mathbf{M}\mathrm{(OH)_x(CO_3)_y\cdot zH_2O+(\frac{2a-x-2y}{4})O_2\to (\frac{x}{2}+z)H_2O+yCO_2+}\mathbf{M}O_a$

从而得到方程：

$\frac{x+y+z}{y}=\frac{2.16/18.016}{0.448/22.4}$

化简得到 $x+z=5y$

<Checkpoint>[推导出未知数关系$x+z=5y$](1)</Checkpoint>

由金属质量可列出第二个方程：

$\frac{10.36}{\mathrm{M}}=\frac{2.16}{18.016(x+y+z)}=\frac{2.16}{18.016\cdot (6y)}=\frac{1}{50.04y}$

试解方程，当$y=2/5$时，有合理解$\mathrm{Pb}$。

<Checkpoint>[通过解不定方程得到$\mathbf{M}$为$\mathrm{Pb}$](1)</Checkpoint>

此时$x+z=5y=2$

$\mathrm{Pb}$常见化合价有+2与+4，在这里显然为+2价，因此由化合价可计算得到$x=\frac{6}{5}$，因此$z=\frac{4}{5}$

故$\mathbf{X}$应为$\mathrm{Pb(OH)_{\frac{6}{5}}(CO_3)_{\frac{2}{5}}\cdot \frac{4}{5}H_2O}$。

<Checkpoint>[$\mathbf{X}$的化学式为$\mathrm{Pb(OH)_{\frac{6}{5}}(CO_3)_{\frac{2}{5}}\cdot \frac{4}{5}H_2O}$](1)</Checkpoint>

因此选择L选项