Weigh out $\mathsf { \pmb M } ( \mathrm { N O } _ { 3 } ) _ { 2 } 9 . 9 3 \ \mathrm { g }$, urea $9 . 0 0 \ \mathrm { g }$, and $\mathrm { P E G-2 0 0 0  }$ $3 \ \mathrm { g }$, dissolve them in ${ 8 0 0 } \mathrm { m L }$ of deionized water, and heat in an oil bath at $1 0 5 ~ ^ { \circ } \mathrm{ C }$. After reacting for $5 \mathrm { h }$, filter under suction. Wash the product three times each with deionized water and anhydrous ethanol solution, then dry at $8 0 ~ ^ { \circ } \mathrm{ C }$ for $\mathrm { 8 h }$ to obtain $6 . 7 0 9 \ \mathrm { g }$ of basic salt $\mathsf { \pmb A }$ with a yield of $8 6 . 5 6 \%$ (based on $\mathsf { M }$). Thermogravimetric analysis of $\mathsf { \pmb A }$ shows an initial weight loss of $2 . 2 \%$, corresponding to $\mathrm { H_2 } \mathrm { O }$, followed by a combined weight loss of $1 1 . 3 \%$ in the second and third steps, accompanied by the release of a colorless and odorless gas, yielding oxide $\mathsf { \pmb B }$.  

By inference, determine which of the following options is correct.
称取 $\mathsf { \pmb M } ( \mathrm { N O } _ { 3 } ) _ { 2 } 9 . 9 3 \ \mathrm { g }$ ，尿素 $9 . 0 0 \ \mathrm { g }$ 及 $\mathrm { P E G-2 0 0 0  }$ $3 \ \mathrm { g }$ 溶于 ${ 8 0 0 } \mathrm { m L }$ 去离子水中，在 $1 0 5 ~ ^ { \circ } \mathrm{ C }$ 条件下油浴加热，反应 $5 \mathrm { h }$ 后抽滤，产品用去离子水和无水乙醇溶液各洗涤3次，并在 $8 0 ~ ^ { \circ } \mathrm{ C }$ 下干燥 $\mathrm { 8 h }$ ，以 $8 6 . 5 6 \%$ 的产率(以$\mathsf { M }$计)得到 $6 . 7 0 9 \ \mathrm { g }$ 碱式盐 $\mathsf { \pmb A }$ 样品。对$\mathsf { \pmb A }$进行热重分析，其第一次失重 $2 . 2 \%$ ，对应 $\mathrm { H_2 } \mathrm { O }$ ，第二三次共失重 $1 1 . 3 \%$ 并放出无色无味气体，得到氧化物$\mathsf { \pmb B }$。

通过推断，判断以下选项哪个是正确的。
The square of the period number of element **M** is less than 20
Basic salt **A** contains more than four elements
The gas released during the second and third weightlessness periods contains water.
The gas composition released during the second and third weightlessness periods consists of more than one component.
The oxidation state of the metal element in oxide **B** differs from that in basic salt **A**.
In the simplest chemical equation for the reaction generating basic salt **A** in this problem, the sum of the stoichiometric coefficients of all reactants is greater than 20.
**M**元素周期数的平方小于20
碱式盐**A**有多于四种元素
第二三次失重放出的气体含有水
第二三次失重放出的气体成分多于一种
氧化物**B**中金属元素价态与碱式盐**A**中不同
本题目中生成碱式盐**A**的反应的最简化学方程式中各反应物化学计量数之和大于20
Let A contain $n$ M, and B be MO, then:

$$
n m ( \pmb { A } ) / \alpha M ( \pmb { \ A } ) = m ( \mathbf { M } ( \mathrm { N O } _ { 3 } ) _ { 2 } ) / M (\mathbf { M }( \mathrm { N O } _ { 3 } ) _ { 2 } )
$$

If $\omega _ { 1 - 1 , 2 , 3 }$ represents the mass fraction of $n$ B in A (i.e., $1 -$ total weight loss), then:

$n m ( \mathbf { A } ) / \alpha M ( \mathbf { A } ) = \omega _ { 1 - 1 , 2 , 3 } n m ( \mathbf { A } ) / \omega _ { 1 - 1 , 2 , 3 } \alpha M ( \mathbf { A } ) = \omega _ { 1 - 1 , 2 , 3 } nm(\mathbf { A })/αM(n\mathbf { M }\mathrm{O}) = \omega _ { 1 -1 , 2 , 3 }m(\mathbf { A })/αM(\mathbf { M }\mathrm{O})$ 

Solving for $M ( \mathbf { \mathbf { M } } ) = 2 0 8 . 5 ~ \mathbf { g } \mathbf { \cdot } \mathbf { m o l } ^ { - 1 }$, checking the table and combining with the knowledge of elements, we know that **M** is Pb, which is a sixth-period element, and the square of the period number is 36, so A is wrong.

<Checkpoint>[**M** is Pb](3)</Checkpoint>

<Checkpoint>[The square of the period number is 36](1)</Checkpoint>

A is a basic salt, obviously $n > 1$. Combining the three thermogravimetric losses of water and colorless and odorless gas, as well as urea in the reactants, it is preliminarily speculated that the composition of A is $a { \mathrm { P b } } ( { \mathrm { O H } } ) _ { 2 } { \cdot } ( n - a ) { \mathrm { P b } } { \mathrm { C O } } _ { 3 }$. Based on this, enumerate the possibilities. When n is greater than 2 and less than 10, only when n=3, a=1 meets the requirements.

Therefore, $a = 1 , \ n = 3$, that is, the composition of A is $\mathrm { P b } _ { 3 } ( \mathrm { O H } ) _ { 2 } ( \mathrm { C O } _ { 3 } ) _ { 2 }$, and the verification meets the second and third weight loss data. It contains four elements, so option B is wrong.

<Checkpoint>[The composition of **A** is $\mathrm { P b } _ { 3 } ( \mathrm { O H } ) _ { 2 } ( \mathrm { C O } _ { 3 } ) _ { 2 }$ ](2)</Checkpoint>

<Checkpoint>[**A** has four elements](1)</Checkpoint>

The relative molecular mass of **A** is 775.616. After the first weight loss, it no longer contains hydrogen, so the second and third weight losses cannot contain water, and can only be carbon dioxide, each time losing 1 molecule of carbon dioxide. The weight loss is due to the decomposition of carbonate, there is no redox reaction, and there is no valence change.

<Checkpoint>[The second and third weight losses cannot contain water](1)</Checkpoint>

<Checkpoint>[The second and third weight losses are both carbon dioxide](1)</Checkpoint>

<Checkpoint>[**A** and **B** have the same valence](2)</Checkpoint>

The last option requires writing the equation:

$\mathrm{3Pb(NO_3)+4CO(NH_2)_2+10H_2O \rightarrow Pb_3(OH)_2(CO_3)_2+6NH_4NO_3+2NH_4HCO_3}$

or

$\mathrm{3Pb(NO_3)+3CO(NH_2)_2+7H_2O \rightarrow Pb_3(OH)_2(CO_3)_2+6NH_4NO_3+CO_2}$

Therefore, the sum of the stoichiometric numbers of each reactant is greater than 20, which is 26 or 21.

<Checkpoint>[In the chemical reaction equation for the formation of basic salt **A**, the sum of the stoichiometric numbers of each reactant is 26 or 21](3)</Checkpoint>
设A中有 $n$ 个M，且B为MO, 则有：

$$
n m ( \pmb { A } ) / \alpha M ( \pmb { \ A } ) = m ( \mathbf { M } ( \mathrm { N O } _ { 3 } ) _ { 2 } ) / M (\mathbf { M }( \mathrm { N O } _ { 3 } ) _ { 2 } )
$$

若用 $\omega _ { 1 - 1 , 2 , 3 }$ 代表 $n$ B占A的质量分数(即 $1 -$ 总失重), 则有：

$n m ( \mathbf { A } ) / \alpha M ( \mathbf { A } ) = \omega _ { 1 - 1 , 2 , 3 } n m ( \mathbf { A } ) / \omega _ { 1 - 1 , 2 , 3 } \alpha M ( \mathbf { A } ) = \omega _ { 1 - 1 , 2 , 3 } nm(\mathbf { A })/αM(n\mathbf { M }\mathrm{O}) = \omega _ { 1 -1 , 2 , 3 }m(\mathbf { A })/αM(\mathbf { M }\mathrm{O})$ 

解得 $M ( \mathbf { \mathbf { M } } ) = 2 0 8 . 5 ~ \mathbf { g } \mathbf { \cdot } \mathbf { m o l } ^ { - 1 }$ ，查表并结合元素知识可知**M**为Pb，为第六周期元素，周期数平方为36，A错。

<Checkpoint>[**M**为Pb](3)</Checkpoint>

<Checkpoint>[周期数平方为36](1)</Checkpoint>

A为碱式盐，显然 $n > 1$ ，结合三次热重分别失去水和无色无味气体，以及反应物中的尿素，初步猜测A的组成为 $a { \mathrm { P b } } ( { \mathrm { O H } } ) _ { 2 } { \cdot } ( n - a ) { \mathrm { P b } } { \mathrm { C O } } _ { 3 }$ ，则以此为基础进行枚举，在n大于2小于10的情况下，仅有n=3时，a=1符合题设

故 $a = 1 , \ n = 3$ ，即A的组成为 $\mathrm { P b } _ { 3 } ( \mathrm { O H } ) _ { 2 } ( \mathrm { C O } _ { 3 } ) _ { 2 }$ ，且检验符合二三次失重数据。含有四种元素，选项B错。

<Checkpoint>[**A**的组成为 $\mathrm { P b } _ { 3 } ( \mathrm { O H } ) _ { 2 } ( \mathrm { C O } _ { 3 } ) _ { 2 }$ ](2)</Checkpoint>

<Checkpoint>[**A**有四种元素](1)</Checkpoint>

**A**的相对分子质量为775.616，第一次失重后已经不含氢元素，故二三次失重不可能含水，只能为二氧化碳，每次失去1分子二氧化碳。失重为碳酸盐分解，没有氧化还原，没有价态变化。

<Checkpoint>[二三次失重不可能含水](1)</Checkpoint>

<Checkpoint>[二三次失重均为二氧化碳](1)</Checkpoint>

<Checkpoint>[**A**与**B**价态相同](2)</Checkpoint>

最后一个选项要求写出方程式：

$\mathrm{3Pb(NO_3)+4CO(NH_2)_2+10H_2O \rightarrow Pb_3(OH)_2(CO_3)_2+6NH_4NO_3+2NH_4HCO_3}$

或

$\mathrm{3Pb(NO_3)+3CO(NH_2)_2+7H_2O \rightarrow Pb_3(OH)_2(CO_3)_2+6NH_4NO_3+CO_2}$

故各反应物的化学计量数之和大于20，为26或21

<Checkpoint>[生成碱式盐**A**的化学反应方程式中，各反应物化学计量数之和为26或21](3)</Checkpoint>