The following chemical processes are described:  
1. In early industry, sodium amide reacts with carbon to produce a highly toxic substance.  
2. Orpiment reacts with a hydrochloric acid solution of stannous chloride to yield a substance that can be used as an insecticide or antidote.  
3. When excess sodium cyanide is added to a mercury-containing container, the mercury completely dissolves after some time.  
4. Dry $\mathrm{NI_3\cdot NH_3}$ is highly prone to explosive decomposition, producing two iodine-containing substances.  
5. Dry iron powder reacts with excess nitric acid of varying concentrations, yielding products that may be **A, B, C, D, E**, with nitrogen mass percentages of approximately $35\%, 100\%, 64\%, 47\%, 31\%$, respectively.  

Write and balance the chemical equations for these processes (with coefficients in the simplest integer ratio). Which of the following statements is correct?
有下列化学过程：
1. 早期工业用氨基钠和碳反应制备一种剧毒物质。
2. 雌黄与二氯化锡的盐酸溶液反应得到可用于杀虫解毒的物质。
3. 将过量氰化钠加入含汞的容器中，一段时间后汞完全溶解。
4. 干燥的$\mathrm{NI_3\cdot NH_3}$极易发生爆炸过程，产物有两种含碘的物质。
5. 干燥铁粉与过量不同浓度的硝酸反应，产物分别可能为**A,B,C,D,E**，它们的氮元素质量分数依次大约为$35\%,100\%,64\%,47\%,31\%$。

将这些过程的化学方程式写出并配平（系数为最简整数比），下列说法正确的是
The equation for Process 1 does not generate gas.
The sum of the reactant coefficients in the equation for Process 2 is 7.
The sum of the reactant coefficients in the equation for Process 3 is 12.
The sum of the product coefficients in the equation of process 4 is not a multiple of 4.
Process 5 generates the equation for **A**, where the sum of product coefficients is the smallest among the equations for generating **A-E**.
Process 5 generates the equation for **B**, where the sum of product coefficients is the largest among the equations generating **A-E**.
Process 5 generates the equation for **C**, where the sum of reactant coefficients is the largest among all equations producing **A-E**, without exception.
Process 5 generates the equation for **D**, where the sum of reactant coefficients is the smallest among the equations producing **A-E**.
Process 5 generates the equation for **E**, where the sum of the product coefficients is the only odd number among the equations for **A-E**.
None of the above options are correct
过程1的方程式无气体产生
过程2的方程式反应物系数和为7
过程3的方程式反应物系数和为12
过程4的方程式产物系数和不是4的倍数
过程5生成**A**的方程式，产物系数和是生成**A-E**的方程式中最小的
过程5生成**B**的方程式，产物系数和不是生成**A-E**的方程式中最大的
过程5生成**C**的方程式，反应物系数和是生成**A-E**的方程式中最大的，没有之一
过程5生成**D**的方程式，反应物系数和是生成**A-E**的方程式中最小的
过程5生成**E**的方程式，产物系数和是生成**A-E**的方程式中唯一的奇数
以上选项均不正确
Process 1: The highly toxic substance is clearly sodium cyanide, while hydrogen gas is released. Option A is incorrect. The equation is  

$\mathrm{NaNH_2}+\mathrm{C}=\mathrm{NaCN}+\mathrm{H_2}$  

<Checkpoint>[$\mathrm{NaNH_2}+\mathrm{C}=\mathrm{NaCN}+\mathrm{H_2}$](1)</Checkpoint>  

Process 2: For insecticidal detoxification, the substance is clearly realgar $\mathrm{As_4S_4}$. Divalent tin is used for reduction and is itself oxidized to tetravalent tin. The equation is  

$\mathrm{2As_2S_3 + 2SnCl_2 + 4HCl = As_4S_4 + 2H_2S + 2SnCl_4}$  

<Checkpoint>[The insecticidal detoxification substance is realgar $\mathrm{As_4S_4}$](1)</Checkpoint>  

<Checkpoint>[$\mathrm{2As_2S_3 + 2SnCl_2 + 4HCl = As_4S_4 + 2H_2S + 2SnCl_4}$](1)</Checkpoint>  

The sum of the reactant coefficients is 8, so option B is incorrect. If considering the form of $\mathrm{Sn(II)}$ in the hydrochloric acid system as $\mathrm{H_2SnCl_4}$ and $\mathrm{Sn(IV)}$ as $\mathrm{H_2SnCl_6}$, the sum of the equation coefficients still does not equal 7.  

Process 3: Divalent mercury can be coordinated by cyanide to form the stable $\mathrm{Hg(CN)_4^{2-}}$. During the dissolution of mercury, it is oxidized by oxygen in the air. Thus, the chemical equation is written as  

$\mathrm{2Hg+8NaCN+2H_2O+O_2=2Na_2[Hg(CN)_4]+4NaOH}$  

<Checkpoint>[$\mathrm{2Hg+8NaCN+2H_2O+O_2=2Na_2[Hg(CN)_4]+4NaOH}$](1)</Checkpoint>  

The sum of the reactant coefficients is 13, so option C is incorrect.  

Process 4: $\mathrm{NI_3\cdot NH_3}$ undergoes an explosion because the process is a highly intense redox reaction, producing stable species. It can be deduced that nitrogen gas and iodine are formed, while the coordinated ammonia combines with the iodide ions generated from disproportionation. Thus, the equation is:  

$\mathrm{8NI_3\cdot NH_3=5N_2+6NH_4I+9I_2}$  

<Checkpoint>[$\mathrm{8NI_3\cdot NH_3=5N_2+6NH_4I+9I_2}$](1)</Checkpoint>  

The sum of the product coefficients is 20, which is a multiple of 4, so option D is incorrect.  

Process 5: First, deduce **A-E**. Clearly, **B** is nitrogen gas. Then, assume the general formula for **A-E** is $N_xH_yO_z$ and infer based on mass fractions. The final results for **A-E** are ammonium nitrate, nitrogen gas, nitrous oxide, nitric oxide, and nitrogen dioxide.  

<Checkpoint>[**A-E** are ammonium nitrate, nitrogen gas, nitrous oxide, nitric oxide, and nitrogen dioxide](1)</Checkpoint>  

Since it is a reaction between iron powder and nitric acid, the product must be iron(III) nitrate. Thus, the following five redox equations are balanced:  

<Checkpoint>[The product must be iron(III) nitrate](0.5)</Checkpoint>  

$\mathrm{8Fe+30HNO_3=8Fe(NO_3)_3+3NH_4NO_3+9H_2O}$  

<Checkpoint>[$\mathrm{8Fe+30HNO_3=8Fe(NO_3)_3+3NH_4NO_3+9H_2O}$](1)</Checkpoint>  

$\mathrm{10Fe+36HNO_3=10Fe(NO_3)_3+3N_2+18H_2O}$  

<Checkpoint>$\mathrm[10Fe+36HNO_3=10Fe(NO_3)_3+3N_2+18H_2O]}(1)</Checkpoint>  

$\mathrm{8Fe+30HNO_3=8Fe(NO_3)_3+3N_2O+15H_2O}$  

<Checkpoint>[$\mathrm{8Fe+30HNO_3=8Fe(NO_3)_3+3N_2O+15H_2O}$](1)</Checkpoint>  

$\mathrm{Fe+4HNO_3=Fe(NO_3)_3+NO+2H_2O}$  

<Checkpoint>[$\mathrm{Fe+4HNO_3=Fe(NO_3)_3+NO+2H_2O}$](1)</Checkpoint>  

$\mathrm{Fe+6HNO_3=Fe(NO_3)_3+3NO_2+3H_2O}$  

<Checkpoint>[$\mathrm{Fe+6HNO_3=Fe(NO_3)_3+3NO_2+3H_2O}$](1)</Checkpoint>  

Based on the sum of the product and reactant coefficients, only option H is correct among options E-I.  

In conclusion, the answer is H.
过程1：剧毒物质很显然为氰化钠，同时放出氢气，选项A错误。方程式为

$\mathrm{NaNH_2}+\mathrm{C}=\mathrm{NaCN}+\mathrm{H_2}$

<Checkpoint>[$\mathrm{NaNH_2}+\mathrm{C}=\mathrm{NaCN}+\mathrm{H_2}$](1)</Checkpoint>

过程2：杀虫解毒，该物质很明显为雄黄$\mathrm{As_4S_4}$，二价锡用于还原，自身被氧化到四价，方程式为

$\mathrm{2As_2S_3 + 2SnCl_2 + 4HCl = As_4S_4 + 2H_2S + 2SnCl_4}$

<Checkpoint>[杀虫解毒的物质为雄黄$\mathrm{As_4S_4}$](1)</Checkpoint>

<Checkpoint>[$\mathrm{2As_2S_3 + 2SnCl_2 + 4HCl = As_4S_4 + 2H_2S + 2SnCl_4}$](1)</Checkpoint>

反应物系数和为8，选项B错误。若考虑盐酸体系里$\mathrm{Sn(II)}$存在形态为$\mathrm{H_2SnCl_4}$，$\mathrm{Sn(IV)}$存在形态为$\mathrm{H_2SnCl_6}$，方程式系数和同样不等于7。

过程3：二价汞可以被氰根配位生成稳定的$\mathrm{Hg(CN)_4^{2-}}$，从而汞的溶解过程中汞会被空气中的氧气氧化，因此写出化学方程式

$\mathrm{2Hg+8NaCN+2H_2O+O_2=2Na_2[Hg(CN)_4]+4NaOH}$

<Checkpoint>[$\mathrm{2Hg+8NaCN+2H_2O+O_2=2Na_2[Hg(CN)_4]+4NaOH}$](1)</Checkpoint>

反应物系数和为13，选项C错误。

过程4：$\mathrm{NI_3\cdot NH_3}$ 发生爆炸, 因为爆炸过程是非常剧烈的氧化还原反应，生成稳定物种，可以推出生成氮气和碘单质，用于配合的氨气会与歧化生成的碘负离子结合。从而写出方程式：

$\mathrm{8NI_3\cdot NH_3=5N_2+6NH_4I+9I_2}$

<Checkpoint>[$\mathrm{8NI_3\cdot NH_3=5N_2+6NH_4I+9I_2}$](1)</Checkpoint>

产物系数和为20，是4的倍数，选项D错误。

过程5：首先推断**A-E**。首先明显**B**为氮气，之后可以设**A-E**的通式为$N_xH_yO_z$，根据质量分数进行推断；最终得到**A-E**分别为硝酸铵，氮气，一氧化二氮，一氧化氮和二氧化氮。

<Checkpoint>[**A-E**分别为硝酸铵，氮气，一氧化二氮，一氧化氮和二氧化氮](1)</Checkpoint>

由于是铁粉与硝酸的反应，生成物一定为硝酸铁，从而配平以下的五个氧化还原方程式：

<Checkpoint>[生成物一定为硝酸铁](0.5)</Checkpoint>

$\mathrm{8Fe+30HNO_3=8Fe(NO_3)_3+3NH_4NO_3+9H_2O}$

<Checkpoint>[$\mathrm{8Fe+30HNO_3=8Fe(NO_3)_3+3NH_4NO_3+9H_2O}$](1)</Checkpoint>

$\mathrm{10Fe+36HNO_3=10Fe(NO_3)_3+3N_2+18H_2O}$

<Checkpoint>$\mathrm[10Fe+36HNO_3=10Fe(NO_3)_3+3N_2+18H_2O]}(1)</Checkpoint>

$\mathrm{8Fe+30HNO_3=8Fe(NO_3)_3+3N_2O+15H_2O}$

<Checkpoint>[$\mathrm{8Fe+30HNO_3=8Fe(NO_3)_3+3N_2O+15H_2O}$](1)</Checkpoint>

$\mathrm{Fe+4HNO_3=Fe(NO_3)_3+NO+2H_2O}$

<Checkpoint>[$\mathrm{Fe+4HNO_3=Fe(NO_3)_3+NO+2H_2O}$](1)</Checkpoint>

$\mathrm{Fe+6HNO_3=Fe(NO_3)_3+3NO_2+3H_2O}$

<Checkpoint>[$\mathrm{Fe+6HNO_3=Fe(NO_3)_3+3NO_2+3H_2O}$](1)</Checkpoint>

根据生成物和产物的系数和，可以判断选项E-I中只有H正确。

综上，答案为H。