The following chemical processes exist:
1. Cobalt chloride desiccant changes from blue to pink in air.
2. $\rm{HMTA}$ reacts with nitric acid to produce $\rm{RDX}$ (cyclotrimethylenetrinitramine).
3. Hydrogen peroxide is added to a $\mathrm{pH=0}$ $\mathrm{TiCl_4}$ solution, and the solution turns red. The titanium-containing cation in the product has a titanium mass fraction of $\rm{54.5\%}$.
4. Acetone reacts with sulfuryl chloride, producing only one small molecule.
5. Potassium permanganate is added dropwise to a ferrous chloride solution, and a fleeting blue color is observed. The product $\rm{Mn^{2+}}$ is not to be written in the equation for this process.

Regarding the sum of the coefficients of all substances in the chemical/ionic equations of the above reactions (ionic equations for processes 3 and 5, chemical equations for processes 1, 2, and 4, coefficients are balanced to the simplest integer ratio), there are a total of five numbers. Which of the following statements about these five numbers is correct:
有下列化学过程：
1. 氯化钴变色硅胶在空气中由蓝色变为粉色。
2. $\rm{HMTA}$与硝酸反应生成$\rm{RDX}$(环三亚甲基三硝胺)。
3. 向$\mathrm{pH=0}$的$\mathrm{TiCl_4}$溶液中加入过氧化氢, 溶液变为红色，产物中的含钛阳离子，钛的质量分数为$\rm{54.5\%}$。
4. 丙酮与磺酰氯反应，只生成一种小分子。
5. 向氯化亚铁溶液中加入滴入高锰酸钾,看到一闪而过的蓝色，要求该过程产物不写出$\rm{Mn^{2+}}$。

关于上述反应的化学/离子方程式（过程3，5为离子方程式，过程1，2，4为化学方程式，系数均配平为最简整数比）的所有物质的系数和，共五个数，关于这五个数下列说法正确的是：
All other options are incorrect
All five numbers are less than 10.
All five numbers are greater than 5.
Among the five numbers, there are no prime numbers.
Among five numbers, there exist two integer power exponents of 2.
Among five numbers, there exist two multiples of 5.
Among five numbers, there is no multiple of 3.
其他选项均不正确
五个数均小于10
五个数均大于5
五个数中不存在质数
五个数中存在两个2的整数幂指数
五个数中存在两个5的倍数
五个数中不存在3的倍数
Process 1: It is obvious that this is the water absorption reaction of anhydrous cobalt chloride. According to chemical common sense, the complex is hexahydrate. The equation is as follows:

$\rm{CoCl_2 + 6H_2O = CoCl_2\cdot 6H_2O}$

<Checkpoint>[$\rm{CoCl_2 + 6H_2O = CoCl_2\cdot 6H_2O}$](1)</Checkpoint>

The sum of coefficients is 8.

Process 2: The chemical formula of $\rm{RDX}$ is $\rm{(CH_2)_3N_3(NO_2)_3}$. Obviously, this reaction is a nitration reaction. Considering the preparation of urotropine, it is known that the excess carbon atoms are converted into formaldehyde, and the nitrogen atoms are converted into ammonium ions. Therefore, the equation can be balanced:

<Checkpoint>[Considering the preparation of urotropine, it is known that the excess carbon atoms are converted into formaldehyde, and the nitrogen atoms are converted into ammonium ions](1)</Checkpoint>

$\rm{(CH_2)_6N_4 + 4HNO_3 = (CH_2)_3N_3(NO_2)_3 + 3HCHO + NH_4NO_3}$

<Checkpoint>[$\rm{(CH_2)_6N_4 + 4HNO_3 = (CH_2)_3N_3(NO_2)_3 + 3HCHO + NH_4NO_3}$](1)</Checkpoint>

The sum of coefficients is 10.

Process 3: According to the knowledge of elemental chemistry, the red color should be the color of the peroxo complex of $\rm{Ti(IV)}$; considering the existence form, combined with the mass fraction, it is known that $\rm{Ti_2O_5^{2+}}$ is generated. Balance the ion equation:

<Checkpoint>[The red color should be the color of the peroxo complex of $\rm{Ti(IV)}$](1)</Checkpoint>

<Checkpoint>[Combined with the mass fraction, it is known that $\rm{Ti_2O_5^{2+}}$ is generated](1)</Checkpoint>

$\rm{2TiO^{2+} + 2H_2O_2 = Ti_2O_5^{2+} + 2H_2O}$

<Checkpoint>[$\rm{2TiO^{2+} + 2H_2O_2 = Ti_2O_5^{2+} + 2H_2O}$](1)</Checkpoint>

The sum of coefficients is 7.

Process 4: Referring to the reaction in organic chemistry, sulfuryl chloride can convert the ketone carbonyl group into a dichloro substituent. Since only one small molecule is produced, α-halogenation is excluded. Therefore, the equation can be written as:

<Checkpoint>[Sulfuryl chloride can convert the ketone carbonyl group into a dichloro substituent](1)</Checkpoint>

<Checkpoint>[Only one small molecule is produced, α-halogenation is excluded](1)</Checkpoint>

$\rm{CO(CH_3)_2 + SO_2Cl_2 = H_3CCCl_2CH_3 + SO_3}$

<Checkpoint>[$\rm{CO(CH_3)_2 + SO_2Cl_2 = H_3CCCl_2CH_3 + SO_3}$](1)</Checkpoint>

The sum of coefficients is 4.

Process 5: According to the knowledge of elemental chemistry, the blue color is the intermediate product of oxidation, $\rm{MnO_4^{3-}}$; thus, the equation is balanced:

<Checkpoint>[The blue color is the intermediate product of oxidation, $\rm{MnO_4^{3-}}$](1)</Checkpoint>

$\rm{2Fe^{2+} + MnO_4^- = 2Fe^{3+} + MnO_4^{3-}}$

<Checkpoint>[$\rm{2Fe^{2+} + MnO_4^- = 2Fe^{3+} + MnO_4^{3-}}$](1)</Checkpoint>

The sum of coefficients is 6.

Therefore, the five numbers are 8, 10, 7, 4, and 6, indicating that only option E is correct.

<Checkpoint>[The five numbers are 8, 10, 7, 4, and 6](1)</Checkpoint>
过程1：很明显为无水氯化钴的吸水反应，根据化学常识可知该配合物为六水合。方程式如下：

$\rm{CoCl_2 + 6H_2O = CoCl_2\cdot 6H_2O}$

<Checkpoint>[$\rm{CoCl_2 + 6H_2O = CoCl_2\cdot 6H_2O}$](1)</Checkpoint>

系数和为8。

过程2：$\rm{RDX}$化学式为$\rm{(CH_2)_3N_3(NO_2)_3}$，很明显该反应为硝化反应，联想到乌洛托品的制备可知多余的碳原子被转化为甲醛，氮原子转化为铵根。因此可配平方程式：

<Checkpoint>[联想到乌洛托品的制备可知多余的碳原子被转化为甲醛，氮原子转化为铵根](1)</Checkpoint>

$\rm{(CH_2)_6N_4 + 4HNO_3 = (CH_2)_3N_3(NO_2)_3 + 3HCHO + NH_4NO_3}$

<Checkpoint>[$\rm{(CH_2)_6N_4 + 4HNO_3 = (CH_2)_3N_3(NO_2)_3 + 3HCHO + NH_4NO_3}$](1)</Checkpoint>

系数和为10。

过程3：根据元素化学知识，红色应为$\rm{Ti(IV)}$的过氧配合物的颜色；考虑存在形式，结合质量分数可知生成了$\rm{Ti_2O_5^{2+}}$。配平离子方程式：

<Checkpoint>[红色应为$\rm{Ti(IV)}$的过氧配合物的颜色](1)</Checkpoint>

<Checkpoint>[结合质量分数可知生成了$\rm{Ti_2O_5^{2+}}$](1)</Checkpoint>

$\rm{2TiO^{2+} + 2H_2O_2 = Ti_2O_5^{2+} + 2H_2O}$

<Checkpoint>[$\rm{2TiO^{2+} + 2H_2O_2 = Ti_2O_5^{2+} + 2H_2O}$](1)</Checkpoint>

系数和为7。

过程4：参考有机化学中的反应，硫酰氯可以转化酮羰基为二氯取代基。由于只生成一种小分子，排除α-卤代。从而可以写出方程式：

<Checkpoint>[硫酰氯可以转化酮羰基为二氯取代基](1)</Checkpoint>

<Checkpoint>[只生成一种小分子，排除α-卤代](1)</Checkpoint>

$\rm{CO(CH_3)_2 + SO_2Cl_2 = H_3CCCl_2CH_3 + SO_3}$

<Checkpoint>[$\rm{CO(CH_3)_2 + SO_2Cl_2 = H_3CCCl_2CH_3 + SO_3}$](1)</Checkpoint>

系数和为4。

过程5：根据元素化学知识，蓝色为氧化的中间产物$\rm{MnO_4^{3-}}$；从而配平方程式：

<Checkpoint>[蓝色为氧化的中间产物$\rm{MnO_4^{3-}}$](1)</Checkpoint>

$\rm{2Fe^{2+} + MnO_4^- = 2Fe^{3+} + MnO_4^{3-}}$

<Checkpoint>[$\rm{2Fe^{2+} + MnO_4^- = 2Fe^{3+} + MnO_4^{3-}}$](1)</Checkpoint>

系数和为6。

因此五个数为8，10，7，4，6，可知只有选项E正确。

<Checkpoint>[五个数为8，10，7，4，6](1)</Checkpoint>