$\mathbf{A}$ is a binary ionic compound containing the anion $\mathbf{X}^－$, which can be synthesized from the corresponding elemental substances under normal conditions via a combination reaction, with a formula weight not exceeding $80 g·mol^{-1}$.  

The quaternary compound $\mathbf{B}$ is a hydrate of a strongly alkaline ionic compound, with mass fractions of C, H, and O being 26.51%, 12.79%, and 52.97%, respectively.  

When solid $\mathbf{B}$ reacts with excess solid $\mathbf{A}$ under vacuum, three products are obtained after extraction and separation with liquid ammonia: a salt $\mathbf{M}$ containing the anion $\mathbf{X}^－$, a strongly alkaline ionic compound $\mathbf{C}$, and a colorless gas $\mathbf{D}$. The anion $\mathbf{X}^－$ is derived from $\mathbf{D}$ by gaining electrons.  

Determine the sum of the stoichiometric coefficients (in their simplest whole-number ratios) for all substances in the overall reaction equation for the formation of compound $\mathbf{M}$ from compounds $\mathbf{A}$ and $\mathbf{B}$.
$\mathbf{A}$是一种包含阴离子$\mathbf{X}^－$的二元离子化合物，可以由相应单质在通常条件下发生化合反应得到，且式量不超过$80 g·mol^{-1}$。四元化合物$\mathbf{B}$是一种强碱性离子化合物的水合物，$\mathbf{B}$中C、H、O的质量分数分别为26.51%、12.79%、52.97%。固体$\mathbf{B}$和过量的固体$\mathbf{A}$在真空下反应，经液氨萃取分离得到三种产物：包含阴离子$\mathbf{X}^－$的盐$\mathbf{M}$，强碱性离子化合物$\mathbf{C}$和无色气体$\mathbf{D}$。$\mathbf{X}^－$为$\mathbf{D}$得电子产生的物种。给出由化合物$\mathbf{A}$和化合物$\mathbf{B}$得到化合物$\mathbf{M}$的总反应方程式中所有物质系数的和（各物质系数均为最简整比）。
63
4
20
11
12
8
None of the above options are correct.
63
4
20
11
12
8
以上选项都不对。
Since compound $\mathbf{A}$ is a binary ionic compound, it is highly likely that its cation and anion correspond to a single element each. Therefore, $\mathbf{X}^－$ contains only one element, and since $\mathbf{X}^－$ is obtained by the electron gain of the colorless gas $\mathbf{D}$, $\mathbf{D}$ must be a colorless gaseous elemental substance, either $O_2$ or $N_2$. As $\mathbf{X}^－$ carries only a single negative charge, $\mathbf{X}^－$ can only be $O_2^－$ or $N_3^－$, and the cation must be an alkali metal ion.

<Checkpoint>[$\mathbf{X}^－$ can only be $O_2^－$ or $N_3^－$](1)</Checkpoint>

<Checkpoint>[The cation must be an alkali metal ion](2)</Checkpoint>

Since compound $\mathbf{A}$ can be formed under normal conditions by the direct combination of elemental substances, and nitrogen gas cannot directly combine with alkali metals other than Li, nor can it produce azide ions even when combined with Li, $\mathbf{D}$ must be oxygen, and $\mathbf{X}^－$ is the superoxide ion. Sodium metal cannot directly combine with oxygen under normal conditions to form a superoxide, whereas potassium can. The atomic weights of rubidium and cesium would violate the condition of a molar mass less than 80 g/mol. Therefore, compound $\mathbf{A}$ must be potassium superoxide $KO_2$, and $\mathbf{D}$ is oxygen.

Nitrogen gas cannot directly combine with alkali metals other than Li, nor can it produce azide ions when combined with Li, so $\mathbf{D}$ must be oxygen, and $\mathbf{X}^－$ is the superoxide ion.

<Checkpoint>[$\mathbf{D}$ is oxygen, $\mathbf{X}^－$ is the superoxide ion](2)</Checkpoint>

Sodium metal cannot directly combine with oxygen under normal conditions to form a superoxide, whereas potassium can. The atomic weights of rubidium and cesium would violate the condition of a molar mass less than 80 g/mol.

<Checkpoint>[Compound $\mathbf{A}$ is potassium superoxide $KO_2$](1)</Checkpoint>

<Checkpoint>[$\mathbf{D}$ is oxygen](1)</Checkpoint>

For compound $\mathbf{B}$, the molar ratio of C, H, and O is determined to be 4:23:6 based on the mass fraction.

<Checkpoint>[The molar ratio of C, H, and O in compound $\mathbf{B}$ is 4:23:6](2)</Checkpoint>

Assuming 6 oxygen atoms, the molar mass of $\mathbf{B}$ is $181.23 g·mol^{-1}$. Subtracting the contributions of C, H, and O leaves 14.01, corresponding exactly to one nitrogen atom. Thus, the chemical formula of $\mathbf{B}$ is $C_4H_{23}O_6N$.

Considering it is a strongly alkaline ionic compound with hydration, the alkalinity must originate from an anion, and its basicity is less than or equal to that of hydroxide. Based on the chemical formula, it is reasonable to infer that the source of alkalinity is hydroxide.

<Checkpoint>[The alkalinity of $\mathbf{B}$ must originate from an anion, and its basicity is less than or equal to that of hydroxide](1)</Checkpoint>

<Checkpoint>[The chemical formula is $C_4H_{23}O_6N$](1)</Checkpoint>

The remaining formula $C_4H_{22}O_{5}N$ corresponds to the cation and water of crystallization. No reasonable solution exists for 1, 2, 3, or 4 water molecules. With 5 water molecules, the cation formula is $C_4H_{12}N$, corresponding to the tetramethylammonium ion. Therefore, the chemical formula of $\mathbf{B}$ is $[Me_4N]^＋[OH]^－·5H_2O$.

<Checkpoint>[The chemical formula of $\mathbf{B}$ is $[Me_4N]^＋[OH]^－·5H_2O$](1)</Checkpoint>

Clearly, based on reactivity, $KO_2$ will react with the water of crystallization in $\mathbf{B}$ to produce $O_2$ and hydroxide ions, so hydroxide must appear among the three products. The tetramethylammonium ion of $\mathbf{B}$ and the potassium ion of $\mathbf{A}$ must each combine with an anion after the reaction. Since the product $\mathbf{M}$ contains the superoxide ion, its cation cannot be potassium (otherwise, no reaction would occur). Therefore, $\mathbf{M}$ is $[Me_4N]^＋[O_2]^－$, and $\mathbf{C}$ must be KOH.

<Checkpoint>[$\mathbf{M}$ is $[Me_4N]^＋[O_2]^－$](1)</Checkpoint>

<Checkpoint>[$\mathbf{C}$ must be KOH](1)</Checkpoint>

Balancing the equation yields:
$22KO_2＋2[Me_4N][OH]·5H_2O＝2[Me_4N][O_2]＋22KOH＋15O_2$

<Checkpoint>[$22KO_2＋2[Me_4N][OH]·5H_2O＝2[Me_4N][O_2]＋22KOH＋15O_2$](2)</Checkpoint>

The sum of the coefficients is 63, so the answer is A.

<Checkpoint>[The sum of the coefficients is 63](1)</Checkpoint>
由于化合物$\mathbf{A}$是二元离子化合物，因此很有可能其阳离子和阴离子分别各自对应了一种元素。因此$\mathbf{X}^－$只包含一种元素，而$\mathbf{X}^－$由无色气体$\mathbf{D}$得电子得到，因此$\mathbf{D}$是一种无色气体单质，为$O_2$或$N_2$。由于$\mathbf{X}^－$只带1个负电荷，因此$\mathbf{X}^－$只能是$O_2^－$或$N_3^－$，且阳离子必为碱金属离子。

<Checkpoint>[$\mathbf{X}^－$只能是$O_2^－$或$N_3^－$](1)</Checkpoint>

<Checkpoint>[阳离子必为碱金属离子](2)</Checkpoint>

由于化合物$\mathbf{A}$可以在通常条件下通过单质化合得到，而氮气与除Li外的碱金属不能直接化合，与Li化合也不能产生叠氮离子，于是$\mathbf{D}$必为氧气，$\mathbf{X}^－$为超氧根。钠单质无法与氧气在通常条件下直接化合为超氧化物，而钾单质可以，铷、铯的原子量会导致式量小于80 g/mol的条件不满足，因此化合物$\mathbf{A}$必为超氧化钾$KO_2$，而$\mathbf{D}$为氧气。

氮气与除Li外的碱金属不能直接化合，与Li化合也不能产生叠氮离子，于是$\mathbf{D}$必为氧气，$\mathbf{X}^－$为超氧根

<Checkpoint>[$\mathbf{D}$为氧气，$\mathbf{X}^－$为超氧根](2)</Checkpoint>

钠单质无法与氧气在通常条件下直接化合为超氧化物，而钾单质可以，铷、铯的原子量会导致式量小于80 g/mol的条件不满足

<Checkpoint>[化合物$\mathbf{A}$为超氧化钾$KO_2$](1)</Checkpoint>

<Checkpoint>[$\mathbf{D}$为氧气](1)</Checkpoint>

对于化合物$\mathbf{B}$，通过质量分数求出其C、H、O的摩尔比为4：23：6。

<Checkpoint>[化合物$\mathbf{B}$的C、H、O的摩尔比为4：23：6](2)</Checkpoint>

假设有6个O原子，则$\mathbf{B}$摩尔质量为$181.23 g·mol^{-1}$，扣去C、H、O剩余14.01，恰好对应一个氮原子，故$\mathbf{B}$化学式为$C_4H_{23}O_6N$。

考虑为强碱性离子化合物，且存在水合现象，则碱性来源必为阴离子，且其碱性小于等于氢氧根。联系化学式，容易推知碱性来源应为氢氧根。

<Checkpoint>[$\mathbf{B}$碱性来源必为阴离子，且其碱性小于等于氢氧根](1)</Checkpoint>

<Checkpoint>[化学式为$C_4H_{23}O_6N$](1)</Checkpoint>

剩余化学式$C_4H_{22}O_{5}N$对应阳离子和结晶水，当结晶水数量为1、2、3、4时均无合理解，当结晶水数量为5时，阳离子化学式为$C_4H_{12}N$，对应于四甲基铵离子。因此$\mathbf{B}$化学式为$[Me_4N]^＋[OH]^－·5H_2O$。

<Checkpoint>[$\mathbf{B}$化学式为$[Me_4N]^＋[OH]^－·5H_2O$](1)</Checkpoint>

显然，根据反应性，$KO_2$将与$\mathbf{B}$中的结晶水反应生成$O_2$并产生氢氧根，因此氢氧根必然出现在三种产物中；而$\mathbf{B}$的阳离子四甲基铵离子和$\mathbf{A}$中的钾离子在反应后必然需要分别结合一种阴离子；而产物$\mathbf{M}$包含了超氧根，因此其阳离子不能为钾离子（否则就没有发生反应了），因此$\mathbf{M}$为$[Me_4N]^＋[O_2]^－$，而$\mathbf{C}$必为KOH。

<Checkpoint>[$\mathbf{M}$为$[Me_4N]^＋[O_2]^－$](1)</Checkpoint>

<Checkpoint>[$\mathbf{C}$必为KOH](1)</Checkpoint>

配平方程式得：
$22KO_2＋2[Me_4N][OH]·5H_2O＝2[Me_4N][O_2]＋22KOH＋15O_2$

<Checkpoint>[$22KO_2＋2[Me_4N][OH]·5H_2O＝2[Me_4N][O_2]＋22KOH＋15O_2$](2)</Checkpoint>

其系数和为63，选A。

<Checkpoint>[系数和为63](1)</Checkpoint>