People have discovered that a certain substance $\mathbf{M}$ can be prepared through the following reactions:
$\begin{gathered}
\mathbf{A}(\mathrm{l})+3 \mathbf{B}(\mathrm{s})=\mathbf{M}(\mathrm{s})+3 \mathbf{C}(\mathrm{s})+4 \mathbf{D}(\mathrm{g}) \\
\mathbf{A}(\mathrm{l})+\mathbf{E}(\mathrm{s})=\mathbf{M}(\mathrm{s})+\mathbf{C}(\mathrm{s})+2 \mathbf{F}(\mathrm{g}) \\
2 \mathbf{G}(\mathrm{s})+\mathbf{H}(\mathrm{s})=2 \mathbf{M}(\mathrm{s})+\mathbf{l}(\mathrm{s})+4 \mathbf{J}(\mathrm{g}) \\
5 \mathbf{G}(\mathrm{s})+2 \mathbf{H}(\mathrm{s})+\mathbf{B}(\mathrm{s})=4 \mathbf{M}(\mathrm{s})+2 \mathbf{I}(\mathrm{s})+10 \mathbf{J}(\mathrm{g})+\mathbf{K}(\mathrm{s})+2 \mathbf{D}(\mathrm{g})
\end{gathered}$

$\mathbf{M}$ is composed of two short-period elements $\mathbf{X}$ and $\mathbf{Y}$ (the atomic number of $\mathbf{X}$ is less than that of $\mathbf{Y}$) and does not contain hydrogen. In $\mathbf{A}$, $\omega(\mathrm{Br})=95.7 \%$, $\mathbf{B}$ is a sodium salt that is relatively stable at room temperature but decomposes into a large amount of gas upon heating. $\mathbf{E}$ is slightly more stable than $\mathbf{B}$, where $\omega(\mathrm{Na})=58.9 \%$. $\mathbf{G}$ decomposes upon heating to release an acidic gas and a basic gas, and $\mathbf{G}$ contains a coordinate covalent bond. $\mathbf{I}$ is a Mg-containing salt, $\omega(\mathrm{Mg})=25.5 \%$.

The following statements about the 12 substances $\bf{A-K,M}$ are correct:
人们发现了通过如下反应可以制得某物质 $\mathbf{M}$ :
$\begin{gathered}
\mathbf{A}(\mathrm{l})+3 \mathbf{B}(\mathrm{s})=\mathbf{M}(\mathrm{s})+3 \mathbf{C}(\mathrm{s})+4 \mathbf{D}(\mathrm{g}) \\
\mathbf{A}(\mathrm{l})+\mathbf{E}(\mathrm{s})=\mathbf{M}(\mathrm{s})+\mathbf{C}(\mathrm{s})+2 \mathbf{F}(\mathrm{g}) \\
2 \mathbf{G}(\mathrm{s})+\mathbf{H}(\mathrm{s})=2 \mathbf{M}(\mathrm{s})+\mathbf{l}(\mathrm{s})+4 \mathbf{J}(\mathrm{g}) \\
5 \mathbf{G}(\mathrm{s})+2 \mathbf{H}(\mathrm{s})+\mathbf{B}(\mathrm{s})=4 \mathbf{M}(\mathrm{s})+2 \mathbf{I}(\mathrm{s})+10 \mathbf{J}(\mathrm{g})+\mathbf{K}(\mathrm{s})+2 \mathbf{D}(\mathrm{g})
\end{gathered}$

$\mathbf{M}$由两种短周期元素$\mathbf{X}$、$\mathbf{Y}$($\mathbf{X}$的原子序数小于$\mathbf{Y}$)组成且不含氢。$\mathbf{A}$中 $\omega(\mathrm{Br})=95.7 \%$, $\mathbf{B}$ 是一种在常温下较稳定但加热分解出大量气体的钠盐, $\mathbf{E}$ 相对于 $\mathbf{B}$ 稳定些许, 其中 $\omega(\mathrm{Na})=58.9 \%$, $\mathbf{G}$受热分解出一种酸性气体和一种碱性气体且 $\mathbf{G}$ 中存在配位键。$\mathbf{I}$ 为含 $\mathrm{Mg}$ 的盐, $\omega(\mathrm{Mg})=25.5 \%$。

下列关于$\bf{A-K,M}$这12种物质，说法正确的是：
All other options are incorrect
12 kinds of substances contain a total of 8 kinds of elements.
12 kinds of substances contain a total of 9 kinds of elements.
Among 12 substances, there are 9 binary compounds.
Of 12 substances, there are 3 elements.
Of the 12 substances, 4 are gases at room temperature and pressure.
Of the 12 substances, three have a molecular weight exceeding 100.
Of the 12 substances, 3 have a molecular weight below 30.
其他选项均不正确
12种物质一共含有8种元素
12种物质一共含有9种元素
12种物质中，二元化合物有9种
12种物质中，单质有3种
12种物质中，常温常压下为气体的有4种
12种物质中，分子量超过100的有3种
12种物质中，分子量低于30的有3种
The main body starts from $\mathbf{A}$ with $\omega(\mathrm{Br})=95.7 \%$:

Since the mass fraction is large, $\mathbf{A}$ is likely to contain multiple $\rm{Br}$. Assume it contains n:

If n is 1, the remaining molecular weight $=79.9/0.957-79.9=3.6$, no reasonable element.

If n is 2, the remaining molecular weight $=(2\times79.9)/0.957-2\times79.9=7.2$, no reasonable element.

If n is 3, the remaining molecular weight $=(3\times79.9)/0.957-3\times79.9=10.8$, which is element $\rm{B}$.

If n is 4, the remaining molecular weight $=(3\times79.9)/0.957-4\times79.9=14.3$, no reasonable element.

If n is 5, the remaining molecular weight $=(5\times79.9)/0.957-5\times79.9=18$, no reasonable element. A larger n is unlikely to exist.

Therefore, $\mathbf{A}$ is $\rm{BBr_3}$;

<Checkpoint>[Infer $\mathbf{A}$ is $\rm{BBr_3}$ based on mass fraction](2)</Checkpoint>

Then start from $\mathbf{E}$; $\omega(\mathrm{Na})=58.9 \%$, considering that most sodium salts are of the 1:1 or 1:2 type, assume $\mathbf{E}$ contains one $\mathrm{Na}$, the remaining molecular weight is $\rm{23.01/0.589-23.01=16.06}$, which is exactly the molecular weight of $\rm{}NH_2^-$; therefore $\mathbf{E}$ is $\mathrm{NaNH_2}$.

<Checkpoint>[Infer $\mathbf{E}$ is $\mathrm{NaNH_2}$ based on mass fraction](1)</Checkpoint>

$\mathbf{B}$ is less stable than $\mathbf{E}$ and releases gas upon heating, and is likely to contain nitrogen; therefore $\mathbf{B}$ is sodium azide $\mathrm{NaN_3}$, which decomposes into nitrogen gas upon heating.

<Checkpoint>[$\mathbf{B}$ is sodium azide $\mathrm{NaN_3}$](1)</Checkpoint>

$\mathrm{NaNH_2}$ and $\rm{BBr_3}$ in a 1:1 ratio can synthesize $\mathbf{M}$, $\rm{BBr_3}$ and $\mathrm{NaN_3}$ in a 1:3 ratio can also synthesize $\mathbf{M}$; and $\mathbf{M}$ does not contain hydrogen; it is sufficient to determine that elements $\mathbf{X,Y}$ in $\mathbf{M}$ are $\mathrm{B,N}$ respectively, and according to the equation ratio, $\mathbf{M}$ is $\mathrm{BN}$.

<Checkpoint>[Determine that elements $\mathbf{X,Y}$ in $\mathbf{M}$ are $\mathrm{B,N}$ respectively](1)</Checkpoint>

<Checkpoint>[$\mathbf{M}$ is $\mathrm{BN}$ according to the equation ratio](2)</Checkpoint>

From this, the chemical equations for the first two synthesis reactions can be written:

$\rm{BBr_3 + 3NaN_3 = BN + 3NaBr + 4N_2}$

<Checkpoint>[$\rm{BBr_3 + 3NaN_3 = BN + 3NaBr + 4N_2}$](1)</Checkpoint>

$\rm{BBr_3 + NaNH_2 = BN + NaBr + 2HBr}$

<Checkpoint>[$\rm{BBr_3 + NaNH_2 = BN + NaBr + 2HBr}$](1)</Checkpoint>

Therefore $\bf{C}$ is $\rm{NaBr}$, $\bf{D}$ is $\rm{N_2}$, $\bf{F}$ is $\rm{HBr}$.

<Checkpoint>[$\bf{C}$ is $\rm{NaBr}$](0.5)</Checkpoint>

<Checkpoint>[$\bf{D}$ is $\rm{N_2}$](0.5)</Checkpoint>

<Checkpoint>[$\bf{F}$ is $\rm{HBr}$](0.5)</Checkpoint>

$\bf{G}$ can be used to synthesize $\rm{BN}$, so it must contain nitrogen or boron; $\bf{G}$ decomposes to produce both alkaline and acidic gases and has a coordinate bond, and the only alkaline gas containing nitrogen or boron is ammonia; therefore $\bf{G}$ contains $\rm(NH_4^+)$, so $\bf{G}$ can only be $\rm{NH_4Cl}$ or $\rm{NH_4Br}$ or $\rm{NH_4I}$.

<Checkpoint>[$\bf{G}$ contains $\rm(NH_4^+)$](1)</Checkpoint>

According to $\mathbf{I}$ being a $\mathrm{Mg}$ containing salt, $\omega(\mathrm{Mg})=25.5 \%$, assume $\mathbf{I}$ contains only one $\mathrm{Mg}$, the remaining molecular weight is 70.9, which is exactly two chlorine elements; therefore $\mathbf{I}$ is $\rm{MgCl_2}$, and at this time it can be determined that $\bf{G}$ can only be $\rm{NH_4Cl}$.

<Checkpoint>[Infer $\mathbf{I}$ is $\rm{MgCl_2}$ based on mass fraction](1)</Checkpoint>

<Checkpoint>[$\bf{G}$ can only be $\rm{NH_4Cl}$](1)</Checkpoint>

$\bf{G}$ can only be $\rm{NH_4Cl}$, then from the third equation, $\bf{H}$ must contain $\mathrm{Mg，B}$, according to the 2:1 reaction ratio of $\bf{G}$ and $\bf{H}$, $\bf{H}$ is $\rm{MgB_2}$, write the equation:

<Checkpoint>[$\bf{H}$ is $\rm{MgB_2}$ according to the 2:1 reaction ratio of $\bf{G}$ and $\bf{H}$](1)</Checkpoint>

$\rm{2NH_4Cl + MgB_2 = 2BN + MgCl_2 + 4H_2}$

<Checkpoint>[$\rm{2NH_4Cl + MgB_2 = 2BN + MgCl_2 + 4H_2}$](1)</Checkpoint>

Therefore $\bf{J}$ is $\rm{H_2}$.

<Checkpoint>[$\bf{J}$ is $\rm{H_2}$](1)</Checkpoint>

Most of the reactants and products of the fourth equation are already known, try to write the equation:

$\rm{5NH_4Cl + 2MgB_2 + NaN_3 = 4BN + 2MgCl_2 + 10H_2 + \bf{K} + \rm{2N_2}}$

According to the law of conservation of elements, $\bf{K}$ is $\rm{NaCl}$.

<Checkpoint>[$\bf{K}$ is $\rm{NaCl}$ according to the law of conservation of elements.](1)</Checkpoint>

The equation is:

$\rm{5NH_4Cl + 2MgB_2 + NaN_3 = 4BN + 2MgCl_2 + 10H_2 + NaCl + \rm{2N_2}}$

<Checkpoint>[$\rm{5NH_4Cl + 2MgB_2 + NaN_3 = 4BN + 2MgCl_2 + 10H_2 + NaCl + \rm{2N_2}}$](1)</Checkpoint>

Summary: Unknown species $\bf{A-K,M}$ are $\rm{BBr_3,NaN_3,NaBr,N_2,NaNH_2,HBr,NH_4Cl,MgB_2,MgCl_2,H_2,NaCl,BN}$ respectively.

<Checkpoint>[$\bf{A-K,M}$ are $\rm{BBr_3,NaN_3,NaBr,N_2,NaNH_2,HBr,NH_4Cl,MgB_2,MgCl_2,H_2,NaCl,BN}$](1)</Checkpoint>

According to the chemical formulas, only option H is correct.
本体从$\mathbf{A}$中 $\omega(\mathrm{Br})=95.7 \%$入手：

由于质量分数很大，$\mathbf{A}$很可能含有多个$\rm{Br}$。假设含有n个：

若n取1，剩余分子量$=79.9/0.957-79.9=3.6$，无合理元素。

n取2，剩余分子量$=(2\times79.9)/0.957-2\times79.9=7.2$，无合理元素。

n取3，剩余分子量$=(3\times79.9)/0.957-3\times79.9=10.8$， 为$\rm{B}$元素。

n取4，剩余分子量$=(3\times79.9)/0.957-4\times79.9=14.3$，无合理元素。

n取5，剩余分子量$=(5\times79.9)/0.957-5\times79.9=18$，无合理元素。更大的n很难存在。

因此，$\mathbf{A}$为$\rm{BBr_3}$；

<Checkpoint>[根据质量分数推断$\mathbf{A}$为$\rm{BBr_3}$](2)</Checkpoint>

之后从$\mathbf{E}$入手；$\omega(\mathrm{Na})=58.9 \%$，考虑多数钠盐均为1：1或1：2型，假设$\mathbf{E}$含有一个$\mathrm{Na}$，剩余分子量为$\rm{23.01/0.589-23.01=16.06}$，刚好为$\rm{}NH_2^-$的分子量；因此$\mathbf{E}$为$\mathrm{NaNH_2}$。

<Checkpoint>[根据质量分数推断$\mathbf{E}$为$\mathrm{NaNH_2}$](1)</Checkpoint>

$\mathbf{B}$比$\mathbf{E}$不稳定且加热放出气体，且大概率含氮；因此$\mathbf{B}$为叠氮化钠$\mathrm{NaN_3}$，加热分解出氮气。

<Checkpoint>[$\mathbf{B}$为叠氮化钠$\mathrm{NaN_3}$](1)</Checkpoint>

$\mathrm{NaNH_2}$与$\rm{BBr_3}$ 1：1可合成$\mathbf{M}$，$\rm{BBr_3}$与$\mathrm{NaN_3}$ 1：3反应也可合成$\mathbf{M}$；以及$\mathbf{M}$不含氢；足以判断$\mathbf{M}$中$\mathbf{X,Y}$元素分别为$\mathrm{B,N}$，且根据方程式比例可知$\mathbf{M}$为$\mathrm{BN}$。

<Checkpoint>[判断$\mathbf{M}$中$\mathbf{X,Y}$元素分别为$\mathrm{B,N}$](1)</Checkpoint>

<Checkpoint>[根据方程式比例可知$\mathbf{M}$为$\mathrm{BN}$](2)</Checkpoint>

由此可以书写前两个合成反应的化学方程式：

$\rm{BBr_3 + 3NaN_3 = BN + 3NaBr + 4N_2}$

<Checkpoint>[$\rm{BBr_3 + 3NaN_3 = BN + 3NaBr + 4N_2}$](1)</Checkpoint>

$\rm{BBr_3 + NaNH_2 = BN + NaBr + 2HBr}$

<Checkpoint>[$\rm{BBr_3 + NaNH_2 = BN + NaBr + 2HBr}$](1)</Checkpoint>

因此$\bf{C}$为$\rm{NaBr}$，$\bf{D}$为$\rm{N_2}$，$\bf{F}$为$\rm{HBr}$。

<Checkpoint>[$\bf{C}$为$\rm{NaBr}$](0.5)</Checkpoint>

<Checkpoint>[$\bf{D}$为$\rm{N_2}$](0.5)</Checkpoint>

<Checkpoint>[$\bf{F}$为$\rm{HBr}$](0.5)</Checkpoint>

$\bf{G}$可用于合成$\rm{BN}$，因此肯定含有氮元素或硼元素；$\bf{G}$分解同时生成碱性气体和酸性气体且有配位键，含氮元素或硼元素的碱性气体只有氨气；因此$\bf{G}$含有$\rm(NH_4^+)$,从而$\bf{G}$只能为$\rm{NH_4Cl}$或者$\rm{NH_4Br}$或者$\rm{NH_4I}$。

<Checkpoint>[$\bf{G}$含有$\rm(NH_4^+)$](1)</Checkpoint>

根据$\mathbf{I}$ 为含 $\mathrm{Mg}$ 的盐, $\omega(\mathrm{Mg})=25.5 \%$可知，假设$\mathbf{I}$只含有一个$\mathrm{Mg}$，剩余分子量为70.9，刚好为两个氯元素；因此$\mathbf{I}$为$\rm{MgCl_2}$，且此时能确定$\bf{G}$只能为$\rm{NH_4Cl}$。

<Checkpoint>[质量分数推断$\mathbf{I}$为$\rm{MgCl_2}$](1)</Checkpoint>

<Checkpoint>[$\bf{G}$只能为$\rm{NH_4Cl}$](1)</Checkpoint>

$\bf{G}$只能为$\rm{NH_4Cl}$，那么由第三个方程式可知$\bf{H}$一定含$\mathrm{Mg，B}$，根据$\bf{G}$与$\bf{H}$的2：1反应比例可知$\bf{H}$为$\rm{MgB_2}$，写出方程式：

<Checkpoint>[根据$\bf{G}$与$\bf{H}$的2：1反应比例可知$\bf{H}$为$\rm{MgB_2}$](1)</Checkpoint>

$\rm{2NH_4Cl + MgB_2 = 2BN + MgCl_2 + 4H_2}$

<Checkpoint>[$\rm{2NH_4Cl + MgB_2 = 2BN + MgCl_2 + 4H_2}$](1)</Checkpoint>

因此$\bf{J}$为$\rm{H_2}$。

<Checkpoint>[$\bf{J}$为$\rm{H_2}$](1)</Checkpoint>

第四个方程式的反应物和产物大部分都已经知晓，尝试写出方程式：

$\rm{5NH_4Cl + 2MgB_2 + NaN_3 = 4BN + 2MgCl_2 + 10H_2 + \bf{K} + \rm{2N_2}}$

根据元素守恒得知$\bf{K}$为$\rm{NaCl}$。

<Checkpoint>[根据元素守恒得知$\bf{K}$为$\rm{NaCl}$。](1)</Checkpoint>

方程式为：

$\rm{5NH_4Cl + 2MgB_2 + NaN_3 = 4BN + 2MgCl_2 + 10H_2 + NaCl + \rm{2N_2}}$

<Checkpoint>[$\rm{5NH_4Cl + 2MgB_2 + NaN_3 = 4BN + 2MgCl_2 + 10H_2 + NaCl + \rm{2N_2}}$](1)</Checkpoint>

总结：未知物种$\bf{A-K,M}$分别为$\rm{BBr_3,NaN_3,NaBr,N_2,NaNH_2,HBr,NH_4Cl,MgB_2,MgCl_2,H_2,NaCl,BN}$。

<Checkpoint>[$\bf{A-K,M}$分别为$\rm{BBr_3,NaN_3,NaBr,N_2,NaNH_2,HBr,NH_4Cl,MgB_2,MgCl_2,H_2,NaCl,BN}$](1)</Checkpoint>

根据化学式可知只有选项H正确。