1. A highly toxic substance is obtained by the fusion reaction of calcium cyanamide, coke, and sodium carbonate, followed by water leaching.

2. When realgar reacts with an excess of concentrated nitric acid, a pale yellow precipitate is obtained.

3. Cobalt(III) nitrate undergoes thermal decomposition to yield cobalt(II,III) oxide.

4. Rhenium hexafluoride undergoes disproportionation in water.

5. $\mathrm{Ga_2(OH)_2Cl_2\cdot 2H_2O}$ dissolves in $\mathrm{NaOH}$ solution, releasing gas.

Let the sums of the simplest integer ratios of the balanced chemical equations for the above five reaction processes be denoted as a, b, c, d, and e, respectively. Then, which of the following statements is correct?
1.由氰胺化钙、焦炭和碳酸钠熔融反应，再用水浸取得到一种剧毒性物质

2.雄黄与足量浓硝酸反应，得到淡黄色沉淀

3.硝酸钴(III)热分解得到四氧化三钴

4.六氟化铼在水中发生歧化反应

5.$\mathrm{Ga_2(OH)_2Cl_2\cdot 2H_2O}$溶于$\mathrm{NaOH}$溶液，放出气体

分别记上述五个反应过程配平后化学方程的最简整数比之和为a、b、c、d、e，则下列说法中正确的是
There does not exist a number greater than 50.
There are 2 numbers greater than 30.
There are 2 numbers less than 10.
There are 3 odd numbers
There are 2 prime numbers
All of the above statements are incorrect
不存在大于50的
大于30的有2个
小于10的有2个
奇数的有3个
质数的有2个
以上说法都错误
For Reaction 1, the structural formula of calcium cyanamide is $\mathrm{CaCN_2}$. Considering the "highly toxic substance" mentioned in the question stem, it can only be $\mathrm{NaCN}$, and the destination of $\mathrm{Ca}$ is $\mathrm{CaCO_3}$. The equation is: $\mathrm{CaCN_2+Na_2CO_3+C=CaCO_3+2NaCN}$, with a sum of 6.  

<Checkpoint>[Reaction 1: $\mathrm{CaCN_2+Na_2CO_3+C=CaCO_3+2NaCN}$](1)</Checkpoint>  

<Checkpoint>[Sum of coefficients for Reaction 1 is 6](0.5)</Checkpoint>  

For Reaction 2, the "pale yellow precipitate" mentioned in the question stem should be $\mathrm{S}$. Since concentrated nitric acid is added, the product should be the highest oxidation state of $\mathrm{As}$, and the concentrated nitric acid is reduced to $\mathrm{NO_2}$. The equation is: $\mathrm{As_4S_4+20HNO_3=4H_3AsO_4+4S+20NO_2+4H_2O}$, with a sum of 53.  

<Checkpoint>[Reaction 2: $\mathrm{As_4S_4+20HNO_3=4H_3AsO_4+4S+20NO_2+4H_2O}$](1)</Checkpoint>  

<Checkpoint>[Sum of coefficients for Reaction 2 is 53](0.5)</Checkpoint>  

For Reaction 3, it can be observed that the decomposition involves redox, so $\mathrm{N}$ should exist as $\mathrm{NO_2}$. The equation is: $\mathrm{6Co(NO_3)_3=2Co_3O_4+18NO_2+5O_2}$, with a sum of 31.  

<Checkpoint>[Reaction 3: $\mathrm{6Co(NO_3)_3=2Co_3O_4+18NO_2+5O_2}$](1)</Checkpoint>  

<Checkpoint>[Sum of coefficients for Reaction 3 is 31](0.5)</Checkpoint>  

For Reaction 4, disproportionation generally forms adjacent oxidation states. Based on elemental properties, $\mathrm{+6}$-valent $\mathrm{Re}$ disproportionates to $\mathrm{+7}$ and $\mathrm{+4}$. The equation is: $\mathrm{3ReF_6+10H_2O=2HReO_4+ReO_2+18HF}$, with a sum of 34.  

<Checkpoint>[Reaction 4: $\mathrm{3ReF_6+10H_2O=2HReO_4+ReO_2+18HF}$](1)</Checkpoint>  

<Checkpoint>[Sum of coefficients for Reaction 4 is 34](0.5)</Checkpoint>  

For Reaction 5, $\mathrm{+2}$-valent $\mathrm{Ga}$ is strongly reducing and forms $\mathrm{H_2}$. The equation is: $\mathrm{Ga_2(OH)_2Cl_2\cdot 2H_2O+4NaOH=2NaGa(OH)_4+2NaCl+H_2}$, with a sum of 10.  

<Checkpoint>[Reaction 5: $\mathrm{Ga_2(OH)_2Cl_2\cdot 2H_2O+4NaOH=2NaGa(OH)_4+2NaCl+H_2}$](1)</Checkpoint>  

<Checkpoint>[Sum of coefficients for Reaction 5 is 10](0.5)</Checkpoint>  

The sum of coefficients for Reaction 2 is greater than 50, so A is incorrect. The sums for Reactions 2, 3, and 4 are greater than 30, so B is incorrect. Only Reaction 1 has a sum of coefficients less than 10, so C is incorrect. Only Reactions 2 and 3 have sums that are odd, so D is incorrect. The sums for Reactions 2 and 3 are prime numbers, so E is correct.
对于反应1，氰胺化钙的结构式为$\mathrm{CaCN_2}$，考虑到题干中指出“剧毒性物质”，只能为$\mathrm{NaCN}$，则$\mathrm{Ca}$的去向为$\mathrm{CaCO_3}$。方程式为：$\mathrm{CaCN_2+Na_2CO_3+C=CaCO_3+2NaCN}$，和为6。

<Checkpoint>[反应1：$\mathrm{CaCN_2+Na_2CO_3+C=CaCO_3+2NaCN}$](1)</Checkpoint>

<Checkpoint>[反应1系数和为6](0.5)</Checkpoint>

对于反应2，题干中指出“淡黄色沉淀”，应当为$\mathrm{S}$。题干中加入浓硝酸，则产物应当为$\mathrm{As}$的最高氧化态，浓硝酸被还原为$\mathrm{NO_2}$。方程式为：$\mathrm{As_4S_4+20HNO_3=4H_3AsO_4+4S+20NO_2+4H_2O}$，和为53。

<Checkpoint>[反应2：$\mathrm{As_4S_4+20HNO_3=4H_3AsO_4+4S+20NO_2+4H_2O}$](1)</Checkpoint>

<Checkpoint>[反应2系数和为53](0.5)</Checkpoint>

对于反应3，可以看出分解过程发生了氧化还原，则$\mathrm{N}$应当以$\mathrm{NO_2}$存在。方程式为：$\mathrm{6Co(NO_3)_3=2Co_3O_4+18NO_2+5O_2}$，和为31。

<Checkpoint>[反应3：$\mathrm{6Co(NO_3)_3=2Co_3O_4+18NO_2+5O_2}$](1)</Checkpoint>

<Checkpoint>[反应3系数和为31](0.5)</Checkpoint>

对于反应4，歧化一般形成相邻价态。根据元素性质，$\mathrm{+6}$的$\mathrm{Re}$会歧化为$\mathrm{+7}$和$\mathrm{+4}$，方程式为：$\mathrm{3ReF_6+10H_2O=2HReO_4+ReO_2+18HF}$，和为34。

<Checkpoint>[反应4：$\mathrm{3ReF_6+10H_2O=2HReO_4+ReO_2+18HF}$](1)</Checkpoint>

<Checkpoint>[反应4系数和为34](0.5)</Checkpoint>

对于反应5，$\mathrm{+2}$价的$\mathrm{Ga}$具有强还原性，会形成$\mathrm{H_2}$。方程式为：$\mathrm{Ga_2(OH)_2Cl_2\cdot 2H_2O+4NaOH=2NaGa(OH)_4+2NaCl+H_2}$，和为10。

<Checkpoint>[反应5：$\mathrm{Ga_2(OH)_2Cl_2\cdot 2H_2O+4NaOH=2NaGa(OH)_4+2NaCl+H_2}$](1)</Checkpoint>

<Checkpoint>[反应5系数和为10](0.5)</Checkpoint>

反应2系数和大于50，A错；反应2、3、4系数和大于30，B错；只有反应1系数和小于10，C错；只有反应2、3系数和为奇数，D错；反应2、3系数和为质数，E正确。