Heating the hydrated salt $\mathbf A$ of an active main group metal $\mathbf M$ under an air atmosphere begins to decompose at 471.2 K, forming compound $\mathbf B$ with a mass loss of 5.39% (theoretical mass fraction, same below; relative atomic masses are retained to two decimal places). Continued heating to 660.2 K produces compound $\mathbf C$ with a mass loss of 34.74%. Finally, decomposition begins at 968.7 K to form compound $\mathbf D$, releasing a colorless and odorless gas $\mathbf X$, with a mass loss of 26.32%. All mass loss percentages are based on $\mathbf A$. Regarding $\mathbf A$~$\mathbf D$, which of the following statements is correct:
在空气氛下加热含结晶水的活泼主族金属$\mathbf M$的盐$\mathbf A$，在471.2 K开始分解生成化合物$\mathbf B$，失重5.39%（理论质量分数，下同；相对原子质量保留至小数点后两位）；继续加热至660.2 K生成化合物$\mathbf C$，失重34.74%；最终直至968.7 K开始分解生成化合物$\mathbf D$，放出一种无色无味气体$\mathbf X$，失重26.32%。所有失重比例均基于$\mathbf A$。关于$\mathbf A$~$\mathbf D$，下列说法正确的是：
The chemical formula of $\mathbf A$ is $\mathbf B\mathrm{·2H_2O}$.
The chemical formula of $\mathbf A$ is $\mathbf B\mathrm{·H_2O}$
$\mathbf B$ reacts with hydrochloric acid to produce a colorless and odorless acid.
$\mathbf C$ does not produce gas when reacting with hydrochloric acid
$\mathbf D$ does not react with water
$\mathbf D$ reacts with hydrochloric acid to yield a toxic salt.
Passing $\mathbf X$ into an aqueous solution of $\mathbf B$ yields $\mathbf C$
The thermal decomposition of $\mathbf B$ into $\mathbf C$ can be regarded as the release of compound $\mathbf E$, whose ideal point group is $\mathrm{C_{2v}}$.
All other options are incorrect
$\mathbf A$的化学式为$\mathbf B\mathrm{·2H_2O}$
$\mathbf A$的化学式为$\mathbf B\mathrm{·H_2O}$
$\mathbf B$与盐酸反应可以得到一种无色无味的酸
$\mathbf C$与盐酸反应不产生气体
$\mathbf D$不与水发生反应
$\mathbf D$与盐酸反应可以得到一种有毒性的盐
向$\mathbf B$的水溶液中通入$\mathbf X$可以得到$\mathbf C$
$\mathbf B$热分解为$\mathbf C$的过程可以视为释放了化合物$\mathbf E$，$\mathbf E$的理想点群为$\mathrm{C_{2v}}$
其它选项均是错误的
Highly reactive main group metals indicate that $\mathbf M$ is an alkali metal or alkaline earth metal.

<Checkpoint>[$\mathbf M$ is an alkali metal or alkaline earth metal](1)</Checkpoint>

Oxides of alkali metals or alkaline earth metals are difficult to decompose, so $\mathbf X$ is carbon dioxide, $\mathbf C$ is carbonate, and $\mathbf D$ is oxide.

<Checkpoint>[$\mathbf X$ is carbon dioxide, $\mathbf C$ is carbonate, and $\mathbf D$ is oxide](1)</Checkpoint>

$\frac{2M + 16}{44.01}=\frac{0.3355}{0.2632}$
or
$\frac{M + 16}{44.01}=\frac{0.3355}{0.2632}$

Solving gives $\mathrm M=40.09$, $\mathbf M \mathrm{= Ca}$.

<Checkpoint>[$\mathbf M \mathrm{= Ca}$](1)</Checkpoint>

A mass loss of 34.74% corresponds to 58.09, which is $\mathrm{C_2H_2O_2}$ or $\mathrm{C_3H_6O}$; the latter is reasonable, meaning the anion is acetate. Consequently, $\mathbf A$ has 0.5 water molecules of crystallization, so $\mathbf A \mathrm{=(CH_3CO_2)_2Ca·0.5H_2O}$, $\mathbf B \mathrm{=(CH_3CO_2)_2Ca}$, $\mathbf C \mathrm{=CaCO_3}$, $\mathbf D \mathrm{=CaO}$.

<Checkpoint>[$\mathbf A \mathrm{=(CH_3CO_2)_2Ca·0.5H_2O}$, $\mathbf B \mathrm{=(CH_3CO_2)_2Ca}$, $\mathbf C \mathrm{=CaCO_3}$, $\mathbf D \mathrm{=CaO}$](2)</Checkpoint>

Options A and B are incorrect; $\mathbf B$ reacts with hydrochloric acid to produce the pungent-smelling acetic acid, so C is incorrect; $\mathbf C$ reacts with hydrochloric acid to produce carbon dioxide, so D is incorrect; $\mathbf D$ reacts with water, so E is incorrect; $\mathbf D$ reacts with hydrochloric acid to produce non-toxic calcium chloride, so F is incorrect; $\mathbf B$ cannot react with $\mathrm {CO_2}$ to form $\mathbf C$, so G is incorrect.

$\mathbf B$ decomposes into $\mathbf C$ and loses acetone, which belongs to the $\mathrm{C_{2v}}$ point group, so H is correct.

<Checkpoint>[$\mathbf E$ is acetone](1)</Checkpoint>
活泼性主族金属说明$\mathbf M$是碱金属或碱土金属

<Checkpoint>[$\mathbf M$为碱金属或碱土金属](1)</Checkpoint>

碱金属或碱土金属氧化物不易分解，故$\mathbf X$为二氧化碳，$\mathbf C$为碳酸盐，$\mathbf D$为氧化物

<Checkpoint>[$\mathbf X$为二氧化碳，$\mathbf C$为碳酸盐，$\mathbf D$为氧化物](1)</Checkpoint>

$\frac{2M + 16}{44.01}=\frac{0.3355}{0.2632}$
或
$\frac{M + 16}{44.01}=\frac{0.3355}{0.2632}$

解得$\mathrm M=40.09$，$\mathbf M \mathrm{= Ca}$

<Checkpoint>[$\mathbf M \mathrm{= Ca}$](1)</Checkpoint>

失重34.74%对应58.09，为$\mathrm{C_2H_2O_2}$或$\mathrm{C_3H_6O}$，后者合理，即酸根为醋酸根；进而推出$\mathbf A$有0.5个结晶水，故$\mathbf A \mathrm{=(CH_3CO_2)_2Ca·0.5H_2O}$，$\mathbf B \mathrm{=(CH_3CO_2)_2Ca}$，$\mathbf C \mathrm{=CaCO_3}$，$\mathbf D \mathrm{=CaO}$

<Checkpoint>[$\mathbf A \mathrm{=(CH_3CO_2)_2Ca·0.5H_2O}$，$\mathbf B \mathrm{=(CH_3CO_2)_2Ca}$，$\mathbf C \mathrm{=CaCO_3}$，$\mathbf D \mathrm{=CaO}$](2)</Checkpoint>

选项A、B错误；$\mathbf B$与盐酸反应得到刺激性气味的醋酸，C错误；$\mathbf C$与盐酸反应产生二氧化碳，D错误；$\mathbf D$与水反应，E错误；$\mathbf D$与盐酸反应得到氯化钙无毒，F错误；$\mathbf B$与$\mathrm {CO_2}$反应不能得到$\mathbf C$，G错误；

$\mathbf B$分解为$\mathbf C$失去了属$\mathrm{C_{2v}}$点群的丙酮，H正确

<Checkpoint>[$\mathbf E$为丙酮](1)</Checkpoint>