The following chemical processes occur:
1. A carbon dioxide-saturated solution that has dissolved siderite is oxidized by oxygen to produce a reddish-brown precipitate.
2. A triatomic gas can be prepared by mixing $\rm CH_4, NH_3, O_2$ at high temperature in the presence of a $\rm Pt/Tr$ catalyst.
3. $\rm ClO_3F$ undergoes solvolysis in liquid ammonia, producing two 1:1 type salts.
4. Phosphorus pentafluoride reacts with $\rm O_2F_2$ at low temperature, producing a 1:1 type salt and releasing a gas.
5. 2,4,6-trinitrophenol and excess bleaching powder solution are added to a calcium hydroxide suspension. An oil-like liquid insoluble in water is obtained in the lower layer of the suspension, and the molecular weight of its components is approximately 164.

For the chemical equations involved in the 5 processes, when the coefficients are balanced to the simplest integer ratio, the following statements are made:

1. The sum of the coefficients of the reactants in process 1 is less than the sum of the coefficients of the products.
2. Two of the four numbers, the sum of the coefficients of the reactants and the sum of the coefficients of the products in process 2, and the sum of the coefficients of the reactants and the sum of the coefficients of the products in process 3, are prime numbers.
3. The sum of the coefficients of the products and the sum of the coefficients of the reactants in process 4 are equal.
4. The sum of the coefficients of the products and the sum of the coefficients of the reactants in process 5 are both prime numbers.

The option containing all the correct statements is:
有下列化学过程：
1. 某溶解了菱铁矿的二氧化碳饱和溶液被氧气氧化生成红褐色沉淀。
2. 高温下在$\rm Pt/Tr$催化剂存在下混合$\rm CH_4, NH_3, O_2$可制备一种三原子气体。
3. $\rm ClO_3F$在液氨中溶剂解，产生两种1：1型盐。
4. 五氟化磷与$\rm O_2F_2$低温反应，生成一种1：1型盐并放出气体。
5. 氢氧化钙悬浊液中加入2，4，6-三硝基苯酚和过量漂白粉溶液，悬浊液下层得到不溶于水的油状液体，其成分的分子量约为164。

5个过程涉及的化学方程式，系数配平为最简整数比时，有下列说法：

1. 过程1的生成物系数和小于产物系数和
2. 过程2的生成物系数和和产物系数和，以及过程3的生成物系数和和产物系数和，这四个数中有两个质数
3. 过程4的产物系数和与反应物系数和相等
4. 过程5的产物系数和与反应物系数和均为质数

包含所有正确说法的选项是：
All other options are incorrect
1
2
3
4
1，2
1，3
1，4
2，3
2，4
3，4
1，2，3
1，2，4
2，3，4
其他选项均不正确
1
2
3
4
1，2
1，3
1，4
2，3
2，4
3，4
1，2，3
1，2，4
2，3，4
Process 1: The existing form of siderite $\rm FeCO_3$ dissolving in carbon dioxide-saturated solution should be $\rm Fe(HCO_3)_2$, which is oxidized to ferric hydroxide by oxygen. Balance the equation:

<Checkpoint>[The existing form of siderite $\rm FeCO_3$ dissolving in carbon dioxide-saturated solution should be $\rm Fe(HCO_3)_2$](1)</Checkpoint>

$\rm 4Fe(HCO_3)_2 + O_2 + 2H_2O = 4Fe(OH)_3 + 8CO_2$

<Checkpoint>[$\rm 4Fe(HCO_3)_2 + O_2 + 2H_2O = 4Fe(OH)_3 + 8CO_2$](1)</Checkpoint>

The sum of reactant coefficients and the sum of product coefficients are 7 and 12, respectively. Statement 1 is incorrect.

Process 2: The triatomic gas is composed of elements $\rm C, H, N, O$, so it should be $\rm HCN$; therefore, balance:

<Checkpoint>[The triatomic gas should be $\rm HCN$](1)</Checkpoint>

$\rm 2CH_4 + 2NH_3 + 3O_2 = 2HCN + 6H_2O$

<Checkpoint>[$\rm 2CH_4 + 2NH_3 + 3O_2 = 2HCN + 6H_2O$](1)</Checkpoint>

The sum of reactant coefficients and the sum of product coefficients are 7 and 8, respectively.

Process 3: For the solvolysis reaction, first consider the autoionization of the solvent to produce $\rm NH_4^+, NH_2^-$. Therefore, the positive ions of the two salts should both be $\rm NH_4^+$, and $\rm NH_2^-$ reacts with $\rm ClO_3F$ to undergo substitution, producing $\rm ClO_3(NH)^-$. Therefore, balance as follows:

<Checkpoint>[The positive ions should both be $\rm NH_4^+$, and $\rm NH_2^-$ reacts with $\rm ClO_3F$ to produce $\rm ClO_3(NH)^-$](1)</Checkpoint>

$\rm ClO_3F + 3NH_3 = NH_4F + NH_4ClO_3(NH)$

<Checkpoint>[$\rm ClO_3F + 3NH_3 = NH_4F + NH_4ClO_3(NH)$](1)</Checkpoint>

The sum of reactant coefficients and the sum of product coefficients are 4 and 2, respectively.

There are two prime numbers in 4, 2, 7, 8. Statement 2 is correct.

Process 4: $\rm O_2F_2$ is a strong oxidant, and the positive ion can only be $\rm O_2^+$; $\rm PF_5$ acts as a Lewis acid to produce $\rm PF_6^-$, and the gas released at the same time should be fluorine gas. Balance as follows:

<Checkpoint>[The positive ion can only be $\rm O_2^+$; $\rm PF_5$ acts as a Lewis acid to produce $\rm PF_6^-$](1)</Checkpoint>

$\rm 2O_2F_2 + 2PF_5 = 2O_2PtF_6 + F_2$

<Checkpoint>[$\rm 2O_2F_2 + 2PF_5 = 2O_2PtF_6 + F_2$](1)</Checkpoint>

The sum of reactant coefficients and the sum of product coefficients are 4 and 3, respectively. Statement 3 is incorrect.

Process 5: The main component of bleaching powder is $\rm Ca(ClO)_2$, in which $\rm Cl(I)$ has strong oxidizing properties and can oxidize the carbon of organic matter to $\rm CO_3^{2-}$; the produced insoluble liquid is obviously an organic chloride. Considering the molecular weight, the liquid component is $\rm CCl_3NO_2$. Therefore, balance the equation:

$\rm 11Ca(ClO)_2 + 2HOC_6H_2(NO_2)_3 = 6CCl_3NO_2 + 6CaCO_3 + 3Ca(OH)_2 + 2CaCl_2$

<Checkpoint>[The main component of bleaching powder is $\rm Ca(ClO)_2$, $\rm Cl(I)$ can oxidize the carbon of organic matter to $\rm CO_3^{2-}$](1)</Checkpoint>

<Checkpoint>[The insoluble liquid is obviously an organic chloride. Considering the molecular weight, the liquid component is $\rm CCl_3NO_2$](1)</Checkpoint>

<Checkpoint>[$\rm 11Ca(ClO)_2 + 2HOC_6H_2(NO_2)_3 = 6CCl_3NO_2 + 6CaCO_3 + 3Ca(OH)_2 + 2CaCl_2$](1)</Checkpoint>

The sum of reactant coefficients and the sum of product coefficients are 13 and 17, respectively. Statement 4 is correct.

Judge the statements based on the equations. Statements 1 and 3 are incorrect, and statements 2 and 4 are correct. Option J is correct.
过程1，菱铁矿$\rm FeCO_3$溶解于二氧化碳饱和溶液中的存在形式应为$\rm Fe(HCO_3)_2$，被氧气氧化为氢氧化铁。配平方程式：

<Checkpoint>[菱铁矿$\rm FeCO_3$溶解于二氧化碳饱和溶液中的存在形式应为$\rm Fe(HCO_3)_2$](1)</Checkpoint>

$\rm 4Fe(HCO_3)_2 + O_2 + 2H_2O = 4Fe(OH)_3 + 8CO_2$

<Checkpoint>[$\rm 4Fe(HCO_3)_2 + O_2 + 2H_2O = 4Fe(OH)_3 + 8CO_2$](1)</Checkpoint>

反应物系数和和产物系数和分别为7，12。说法1错误。

过程2，三原子气体的组成元素有$\rm C,H,N,O$，可想到应当是$\rm HCN$；因此配平：

<Checkpoint>[三原子气体应当是$\rm HCN$](1)</Checkpoint>

$\rm 2CH_4 + 2NH_3 + 3O_2 = 2HCN + 6H_2O$

<Checkpoint>[ $\rm 2CH_4 + 2NH_3 + 3O_2 = 2HCN + 6H_2O$](1)</Checkpoint>

反应物系数和和产物系数和分别为7，8。

过程3，溶剂解反应首先考虑溶剂的自耦电离产生$\rm NH_4^+,NH_2^-$，因此两种盐正离子应当均为$\rm NH_4^+$，$\rm NH_2^-$与$\rm ClO_3F$反应发生取代，产生$\rm ClO_3(NH)^-$。因此配平如下：

<Checkpoint>[正离子应当均为$\rm NH_4^+$，$\rm NH_2^-$与$\rm ClO_3F$反应产生$\rm ClO_3(NH)^-$](1)</Checkpoint>

$\rm ClO_3F + 3NH_3 = NH_4F + NH_4ClO_3(NH)$

<Checkpoint>[$\rm ClO_3F + 3NH_3 = NH_4F + NH_4ClO_3(NH)$](1)</Checkpoint>

反应物系数和和产物系数和分别为4，2。

4，2，7，8中存在两个质数，说法2正确。

过程4，$\rm O_2F_2$为强氧化剂，正离子只能为$\rm O_2^+$；$\rm PF_5$作为lewis酸产生$\rm PF_6^-$，同时放出气体应当为氟气。配平如下：

<Checkpoint>[正离子只能为$\rm O_2^+$；$\rm PF_5$作为lewis酸产生$\rm PF_6^-$](1)</Checkpoint>

$\rm 2O_2F_2 + 2PF_5 = 2O_2PtF_6 + F_2$

<Checkpoint>[$\rm 2O_2F_2 + 2PF_5 = 2O_2PtF_6 + F_2$](1)</Checkpoint>

反应物系数和和产物系数和分别为4，3。说法3错误。

过程5，漂白粉主要成分为$\rm Ca(ClO)_2$，其中$\rm Cl(I)$具有强氧化性，能够氧化有机物的碳至$\rm CO_3^{2-}$；产生的不溶液体明显为有机氯化物，考虑分子量可知该液体成分为$\rm CCl_3NO_2$。因此配平方程：

$\rm 11Ca(ClO)_2 + 2HOC_6H_2(NO_2)_3 = 6CCl_3NO_2 + 6CaCO_3 + 3Ca(OH)_2 + 2CaCl_2$

<Checkpoint>[漂白粉主要成分为$\rm Ca(ClO)_2$，$\rm Cl(I)$能够氧化有机物的碳至$\rm CO_3^{2-}$](1)</Checkpoint>

<Checkpoint>[不溶液体明显为有机氯化物，考虑分子量可知该液体成分为$\rm CCl_3NO_2$](1)</Checkpoint>

<Checkpoint>[$\rm 11Ca(ClO)_2 + 2HOC_6H_2(NO_2)_3 = 6CCl_3NO_2 + 6CaCO_3 + 3Ca(OH)_2 + 2CaCl_2$](1)</Checkpoint>

反应物系数和和产物系数和分别为13，17。说法4正确。

根据方程式判断说法。说法1，3错误，说法2，4正确，选项J正确。