The following chemical processes are described:  

1. In an aqueous solution of ethylenediamine, metallic potassium reacts with $\mathrm{KReO_4}$ to yield a hydride containing $\mathrm{Re~68\%}$.  

2. Ceric ammonium nitrate can be used to determine the content of sodium nitrite.  

3. Aluminum carbide can react with a hydrochloric acid solution of mercury(II) chloride to produce a highly toxic substance.  

4. $\mathrm{CO, B_2H_6}$ react in a 2:1 ratio in tetrahydrofuran to form a highly symmetrical cyclic molecule.  

5. Chloroplatinic acid can accelerate the hydrolysis reaction of sodium borohydride.  

6. Cesium azide can react with $\mathrm{N_5SbF_6}$ in sulfur dioxide solvent, with the expectation of obtaining an $\mathrm{N_8}$ molecule, but only a common gas is produced.  

Regarding the main chemical or ionic equations (with stoichiometric coefficients in their simplest integer ratios) involved in these six processes, which of the following statements is correct?
有下列化学过程：

1. 在乙二胺水溶液中，金属钾与$\mathrm{KReO_4}$反应得到含$\mathrm{Re~68\%}$的氢化物。
2. 利用硝酸铈铵可以测定亚硝酸钠的含量。
3. 碳化铝可以和氯化汞的盐酸溶液反应生成一种剧毒物质。
4. $\mathrm{CO,B_2H_6}$在四氢呋喃中按照2：1反应得到一种对称性极高的环状分子。
5. 氯铂酸可以加速硼氢化钠的水解反应。
6. 叠氮化铯可以与$\mathrm{N_5SbF_6}$在二氧化硫溶剂中反应，预期得到$\mathrm{N_8}$分子但最后只得到了一种常见气体。

关于这六个过程涉及到的主要化学或离子方程式（配平后系数为最简整数比），下列说法正确的是：
In the chemical equation for Process 1, the sum of the coefficients of the reactants is less than 30.
In the chemical equation of process 2, the sum of the coefficients of the products is 10.
In the chemical equation of process 3, the sum of the coefficients of the reactants is less than 20
In process 4, a certain product molecule generated has 12 atoms.
In the ionic equation of Process 5 (the products of this ionic equation need to reflect the coordination state of the elements), the sum of the coefficients of the products is 19.
In the chemical equation of process 6, the sum of the coefficients of the products is an even number.
None of the above options are correct
过程1的化学方程式中，反应物的系数和小于30
过程2的化学方程式中，产物的系数和为10
过程3的化学方程式中，反应物的系数和小于20
过程4生成的某种产物分子有12个原子
过程5的离子方程式中（该离子方程式的产物需要体现元素的配位状态），产物的系数和为19
过程6的化学方程式中，生成物的系数和为偶数
以上选项均不正确
Process 1: Based on the mass fraction, it is easy to deduce that the hydride is $\mathrm{K_2ReH_9}$; it is observed that the oxidation state of $\mathrm{Re}$ does not change; hydrogen in the hydride has a -1 oxidation state, so potassium acts as the reducing agent; thus, the redox equation is balanced:

<Checkpoint>[The hydride is $\mathrm{K_2ReH_9}$](1)</Checkpoint>

$\mathrm{18K + KReO_4 + 13H_2O = K_2ReH_9 + 17KOH}$

<Checkpoint>[$\mathrm{18K + KReO_4 + 13H_2O = K_2ReH_9 + 17KOH}$](1)</Checkpoint>

The sum of the reactant coefficients is 32, so option A is incorrect.

Process 2: Ceric ammonium nitrate acts as the oxidizing agent, oxidizing nitrite to nitrate while being reduced to cerium(III) nitrate. The balanced chemical equation is:

$\mathrm{2(NH_4)_2Ce(NO_3)_6 + NaNO_2 + H_2O = 2Ce(NO_3)_3 + NaNO_3 + 4NH_4NO_3 + 2HNO_3}$

<Checkpoint>[$\mathrm{2(NH_4)_2Ce(NO_3)_6 + NaNO_2 + H_2O = 2Ce(NO_3)_3 + NaNO_3 + 4NH_4NO_3 + 2HNO_3}$](1)</Checkpoint>

The sum of the product coefficients is 9, so option B is incorrect.

Process 3: Aluminum carbide reacts with hydrochloric acid, where the negatively charged carbon bonds with the positively charged hydrogen; the highly toxic product formed should contain $\mathrm{Hg(II)}$, thus likely being dimethylmercury. The balanced equation is:

<Checkpoint>[The highly toxic product is dimethylmercury $\mathrm{Hg(CH_3)_2}$](1)</Checkpoint>

$\mathrm{2Al_4C_3 + 3HgCl_2 + 18HCl = 3Hg(CH_3)_2 + 8AlCl_3}$

<Checkpoint>[$\mathrm{2Al_4C_3 + 3HgCl_2 + 18HCl = 3Hg(CH_3)_2 + 8AlCl_3}$](1)</Checkpoint>

The sum of the reactant coefficients is 22, so option C is incorrect.

Process 4: This is a classic reaction for synthesizing boroxine analogs:

$\mathrm{6CO + 3B_2H_6 = 2B_3O_3(CH_3)_3}$

<Checkpoint>[$\mathrm{6CO + 3B_2H_6 = 2B_3O_3(CH_3)_3}$](1)</Checkpoint>

The product contains 18 atoms, so option D is incorrect.

Process 5: Chloroplatinic acid accelerates the hydrolysis of sodium borohydride, which produces borate ion $\mathrm{B(OH)_4^-}$ and hydrogen gas. With the addition of chloroplatinic acid, hydride ions reduce platinum(IV) to platinum metal. The balanced redox equation is:

<Checkpoint>[Sodium borohydride hydrolysis produces borate ion $\mathrm{B(OH)_4^-}$, and hydride ions reduce platinum(IV) to platinum metal](1)</Checkpoint>

$\mathrm{BH_4^- + 2PtCl_6^{2-} + 8OH^- = 2Pt + 12Cl^- + B(OH)_4^- + 4H_2O}$

<Checkpoint>[$\mathrm{BH_4^- + 2PtCl_6^{2-} + 8OH^- = 2Pt + 12Cl^- + B(OH)_4^- + 4H_2O}$](1)</Checkpoint>

The sum of the product coefficients is 19, so option E is correct.

Process 6: The reaction was expected to produce $\mathrm{N_8}$ from the reaction of azide ion with $\mathrm{N_5^+}$, but the actual common gas produced is nitrogen. Thus, the equation is:

$\mathrm{CsN_3 + N_5SbF_6 = 4N_2 + CsSbF_6}$

<Checkpoint>[$\mathrm{CsN_3 + N_5SbF_6 = 4N_2 + CsSbF_6}$](1)</Checkpoint>

The sum of the product coefficients is 5, an odd number, so option F is incorrect.

In conclusion, option E is correct.
过程1：根据质量分数很容易推断氢化物是$\mathrm{K_2ReH_9}$；发现$\mathrm{Re}$的价态不发生变化；氢化物中氢为-1价，则钾作为还原剂；从而配平氧化还原方程：

<Checkpoint>[氢化物是$\mathrm{K_2ReH_9}$](1)</Checkpoint>

$\mathrm{18K + KReO_4 + 13H_2O = K_2ReH_9 + 17KOH}$

<Checkpoint>[$\mathrm{18K + KReO_4 + 13H_2O = K_2ReH_9 + 17KOH}$](1)</Checkpoint>

反应物系数和为32，选项A错误。

过程2：硝酸铈铵作为氧化剂，氧化亚硝酸根至硝酸根，自身被还原为硝酸铈(III)，配平化学方程式：

$\mathrm{2(NH_4)_2Ce(NO_3)_6 + NaNO_2 + H_2O = 2Ce(NO_3)_3 + NaNO_3 + 4NH_4NO_3 + 2HNO_3}$

<Checkpoint>[$\mathrm{2(NH_4)_2Ce(NO_3)_6 + NaNO_2 + H_2O = 2Ce(NO_3)_3 + NaNO_3 + 4NH_4NO_3 + 2HNO_3}$](1)</Checkpoint>

产物系数和为9，选项B错误。

过程3：碳化铝与盐酸反应，带负电的碳会和带正电的氢结合；生成的剧毒物质应当含有$\mathrm{Hg(II)}$，从而应当为甲基汞。因此配平方程式：

<Checkpoint>[剧毒物质为甲基汞$\mathrm{Hg(CH_3)_2}$](1)</Checkpoint>

$\mathrm{2Al_4C_3 + 3HgCl_2 + 18HCl = 3Hg(CH_3)_2 + 8AlCl_3}$

<Checkpoint>[$\mathrm{2Al_4C_3 + 3HgCl_2 + 18HCl = 3Hg(CH_3)_2 + 8AlCl_3}$](1)</Checkpoint>

反应物系数和为22，选项C错误。

过程4：这是合成硼苯类似物的经典反应：

$\mathrm{6CO + 3B_2H_6 = 2B_3O_3(CH_3)_3}$

<Checkpoint>[$\mathrm{6CO + 3B_2H_6 = 2B_3O_3(CH_3)_3}$](1)</Checkpoint>

产物含有18个原子，选项D错误。

过程5：氯铂酸可以加速硼氢化钠的水解反应，硼氢化钠水解生成硼酸根离子$\mathrm{B(OH)_4^-}$和氢气，加入氯铂酸，氢负离子会去还原铂(IV)生成铂单质，从而配平氧化还原方程式：

<Checkpoint>[硼氢化钠水解生成硼酸根离子$\mathrm{B(OH)_4^-}$，氢负离子还原铂(IV)生成铂单质](1)</Checkpoint>

$\mathrm{BH_4^- + 2PtCl_6^{2-} + 8OH^- = 2Pt + 12Cl^- + B(OH)_4^- + 4H_2O}$

<Checkpoint>[$\mathrm{BH_4^- + 2PtCl_6^{2-} + 8OH^- = 2Pt + 12Cl^- + B(OH)_4^- + 4H_2O}$](1)</Checkpoint>

产物系数和为19，选项E正确。

过程6：反应预期叠氮酸根会和$\mathrm{N_5^+}$反应生成$\mathrm{N_8}$，但实际生成的常见气体应该为氮气，故方程式为

$\mathrm{CsN_3 + N_5SbF_6 = 4N_2 + CsSbF_6}$

<Checkpoint>[$\mathrm{CsN_3 + N_5SbF_6 = 4N_2 + CsSbF_6}$](1)</Checkpoint>

产物系数和为5，为奇数，选项F错误。

综上，选项E正确。