The following chemical processes exist:

1.  Zirconium nitride $(\mathrm{ZrN})$ is a relatively inert solid material, but it can dissolve in hot, concentrated sulfuric acid to form a 1:1 type ionic compound, releasing only a simple gaseous substance during the dissolution process.

2.  Under a nitrogen atmosphere, a mixed solid of $\mathrm{B_2O_3}$ and $\mathrm{CaB_6}$ is co-heated, producing only two binary compounds, one of which can be used as a wear-resistant material.

3.  Treating $\mathrm{Tb_4O_7}$ with $\rm 0.1 ~mol/L$ dilute hydrochloric acid can produce pure $\mathrm{TbO_2}$.

4.  Magnetite is added to an excess of $\mathrm{K_2Cr_2O_7-H_2SO_4}$ solution to dissolve it.

5.  In an alkaline sodium hydroxide solution, nickel sulfide reacts quantitatively with carbon monoxide in a molar ratio of $1: 5$.

Given that the stoichiometric coefficients are balanced as integer ratios, the correct option regarding the chemical equations involved in the above five processes is:
有下列化学过程：

1. 氮化锆 $(\mathrm{ZrN})$ 是一种较惰性的固体材料, 但可溶于热的浓硫酸生成 1：1型离子化合物, 溶解过程中只放出单质气体。

2. 在氮气气氛下, 将 $\mathrm{B_2O_3}$ 和 $\mathrm{CaB_6}$ 的混合固体共热, 仅生成了两种二元化合物, 其中一种可用作耐磨损材料。

3. 用 $\rm 0.1 ~mol/L$ 稀盐酸处理 $\mathrm{Tb_4O_7}$, 可制得纯净的 $\mathrm{TbO_2}$。

4. 将磁铁矿投入过量的 $\mathrm{K_2Cr_2O_7-H_2SO_4}$ 溶液中使之溶解。

5. 在氢氧化钠碱性溶液中, 硫化镍与一氧化碳以摩尔比 $1: 5$ 定量反应。

上述五个过程涉及到的化学方程式，在系数配平为整数比的前提下，下列选项正确的是：
All other options are incorrect
The sum of reactant coefficients for Process 1 is 8.
The reactant coefficients of Process 2 are all less than 10.
The sum of product coefficients in Process 3 is greater than the sum of reactant coefficients in Process 1.
The absolute value of the difference between the sum of reactant coefficients and the sum of product coefficients in Process 4 is a prime number.
The ratio of the sum of reactant coefficients to the sum of product coefficients in Process 5 is a decimal.
其他选项均不正确
过程1的反应物系数和为8
过程2的反应物系数均小于10
过程3的产物系数和大于过程1反应物系数和
过程4的反应物系数和减去产物系数和的差的绝对值为质数。
过程5的反应物系数和除以产物系数和，商为小数。
Process 1: $\rm Zr$ is a congener of $\rm Ti$. Similar to $\rm TiO^{2+}$ in concentrated sulfuric acid, $\rm ZrO^{2+}$ should be generated. Therefore, the product is $\rm ZrOSO_4$, which conforms to a 1:1 type ionic compound. Only elemental gas is released, which can only be hydrogen gas. Balancing the equation:

<Checkpoint>[$\rm Zr$ in concentrated sulfuric acid should generate $\rm ZrO^{2+}$ similar to $\rm TiO^{2+}$](1)</Checkpoint>

$\rm 2ZrN + 3H_2SO_4 + 2H_2O = 2ZrOSO_4 + (NH_4)_2SO_4 + H_2$

<Checkpoint>[$\rm 2ZrN + 3H_2SO_4 + 2H_2O = 2ZrOSO_4 + (NH_4)_2SO_4 + H_2$](1)</Checkpoint>

The sum of the reactant coefficients is 7, so option B is incorrect.

Process 2: Under nitrogen atmosphere, the wear-resistant material is obviously $\rm BN$. According to the law of conservation of elements, the other product can only be $\rm CaO$. Balancing the equation:

<Checkpoint>[The wear-resistant material is obviously $\rm BN$](1)</Checkpoint>

$\rm B_2O_3 + 10N_2 + 3CaB_6 = 3CaO + 20BN$

<Checkpoint>[$\rm B_2O_3 + 10N_2 + 3CaB_6 = 3CaO + 20BN$](1)</Checkpoint>

The reactant coefficients include 10, so option C is incorrect.

Process 3: From the perspective of average oxidation state, $\mathrm{Tb_4O_7}$ obviously undergoes a disproportionation reaction. The oxidation state of $\mathrm{TbO_2}$ increases. As a rare earth element, the stable oxidation state can only decrease to $\rm Tb(III)$, so the product is $\rm TbCl_3$.

From another perspective, $\mathrm{Tb_4O_7}$ can be regarded as $\rm 2TbO_2 + Tb_2O_3$. $\rm TbO_2$ is prepared, so the other molecular product can be regarded as $\rm Tb_2O_3$ dissolved in hydrochloric acid to obtain $\rm TbCl_3$.

Balancing the equation:

<Checkpoint>[The other molecular product is $\rm TbCl_3$](1)</Checkpoint>

$\rm Tb_4O_7 + 6HCl = 2TbCl_3 + 3H_2O + 2TbO_2$

<Checkpoint>[$\rm Tb_4O_7 + 6HCl = 2TbCl_3 + 3H_2O + 2TbO_2$](1)</Checkpoint>

The sum of the product coefficients is 7, so option D is incorrect.

Process 4: Magnetite is $\rm Fe_3O_4$. Acidic potassium dichromate has strong oxidizing properties. After dissolution, $\rm Fe(III)$ is generated, and the reduction product of potassium dichromate is $\rm Cr(III)$. Therefore, balancing the chemical equation:

<Checkpoint>[Magnetite is $\rm Fe_3O_4$](1)</Checkpoint>

<Checkpoint>[Acidic potassium dichromate has strong oxidizing properties, and $\rm Fe(III)$ is generated after dissolution](1)</Checkpoint>

$\rm K_2Cr_2O_7 + 6Fe_3O_4 + 31H_2SO_4 = K_2SO_4 + 9Fe_2(SO_4)_3 + Cr_2(SO_4)_3 + 31H_2O$

<Checkpoint>[$\rm K_2Cr_2O_7 + 6Fe_3O_4 + 31H_2SO_4 = K_2SO_4 + 9Fe_2(SO_4)_3 + Cr_2(SO_4)_3 + 31H_2O$](1)</Checkpoint>

The sum of the reactant coefficients is 38, and the sum of the product coefficients is 42. The difference is -4, which is a composite number, so option E is incorrect.

Process 5: The reaction of nickel with carbon monoxide can be thought of as generating the famous carbonyl compound $\rm Ni(CO)_4$; it is a redox reaction. Since it is a 1:5 reaction, the remaining one molecule of $\rm CO$ acts as a reducing agent and is oxidized to $\rm CO_2$, which is written as $\rm CO_3^{2-}$ under alkaline conditions. Therefore, balancing the equation:

<Checkpoint>[Generating the carbonyl compound $\rm Ni(CO)_4$; it is a redox reaction](1)</Checkpoint>

<Checkpoint>[The remaining one molecule of $\rm CO$ acts as a reducing agent](1)</Checkpoint>

$\rm NiS + 5CO + 4NaOH = Ni(CO)_4 + Na_2CO_3+ Na_2S + 2H_2O$

<Checkpoint>[$\rm NiS + 5CO + 4NaOH = Ni(CO)_4 + Na_2CO_3+ Na_2S + 2H_2O$](1)</Checkpoint>

The sum of the reactant coefficients is 10, and the sum of the product coefficients is 5. The quotient is an integer 2, so option F is incorrect.

Options B-F are all incorrect, so option A is correct.
过程1：$\rm Zr$为$\rm Ti$的同族元素，在浓硫酸中类似$\rm TiO^{2+}$应当生成$\rm ZrO^{2+}$，故产物为$\rm ZrOSO_4$，符合1：1型离子化合物。只放出单质气体，只能放出氢气，配平方程式：

<Checkpoint>[$\rm Zr$在浓硫酸中类似$\rm TiO^{2+}$应当生成$\rm ZrO^{2+}$](1)</Checkpoint>

$\rm 2ZrN + 3H_2SO_4 + 2H_2O = 2ZrOSO_4 + (NH_4)_2SO_4 + H_2$

<Checkpoint>[$\rm 2ZrN + 3H_2SO_4 + 2H_2O = 2ZrOSO_4 + (NH_4)_2SO_4 + H_2$](1)</Checkpoint>

反应物系数和为7，选项B错误。

过程2：氮气条件下，耐磨损材料很明显为$\rm BN$，根据元素守恒另一种产物只能为$\rm CaO$。配平方程式：

<Checkpoint>[耐磨损材料显为$\rm BN$](1)</Checkpoint>

$\rm B_2O_3 + 10N_2 + 3CaB_6 = 3CaO + 20BN$

<Checkpoint>[$\rm B_2O_3 + 10N_2 + 3CaB_6 = 3CaO + 20BN$](1)</Checkpoint>

反应物系数存在10，选项C错误。

过程3：从平均氧化态角度，$\mathrm{Tb_4O_7}$很明显发生了歧化反应，$\mathrm{TbO_2}$氧化态升高，作为稀土元素稳定氧化态只能下降到$\rm Tb(III)$，故产物为$\rm TbCl_3$。

另一种角度，$\mathrm{Tb_4O_7}$可以看作$\rm 2TbO_2 + Tb_2O_3$，制备得到$\rm TbO_2$，故另一分子产物可看作为$\rm Tb_2O_3$溶于盐酸，得到$\rm TbCl_3$

配平方程式：

<Checkpoint>[另一分子产物为$\rm TbCl_3$](1)</Checkpoint>

$\rm Tb_4O_7 + 6HCl = 2TbCl_3 + 3H_2O + 2TbO_2$

<Checkpoint>[$\rm Tb_4O_7 + 6HCl = 2TbCl_3 + 3H_2O + 2TbO_2$](1)</Checkpoint>

产物系数和为7，选项D错误。

过程4：磁铁矿为$\rm Fe_3O_4$，酸性重铬酸钾具有强氧化性，溶解后生成$\rm Fe(III)$，重铬酸钾的还原产物为$\rm Cr(III)$。因此配平化学方程式：

<Checkpoint>[磁铁矿为$\rm Fe_3O_4$](1)</Checkpoint>

<Checkpoint>[酸性重铬酸钾具有强氧化性，溶解后生成$\rm Fe(III)$](1)</Checkpoint>

$\rm K_2Cr_2O_7 + 6Fe_3O_4 + 31H_2SO_4 = K_2SO_4 + 9Fe_2(SO_4)_3 + Cr_2(SO_4)_3 + 31H_2O$

<Checkpoint>[$\rm K_2Cr_2O_7 + 6Fe_3O_4 + 31H_2SO_4 = K_2SO_4 + 9Fe_2(SO_4)_3 + Cr_2(SO_4)_3 + 31H_2O$](1)</Checkpoint>

反应物系数和为38，产物系数和为42，差为4，为合数，选项E错误。

过程5：镍与一氧化碳的反应可以想到生成了著名的羰基化合物$\rm Ni(CO)_4$；为氧化还原反应。由于为1：5反应，剩余一分子$\rm CO$充当还原剂被氧化为$\rm CO_2$，碱性条件写为$\rm CO_3^{2-}$。因此配平方程式：

<Checkpoint>[生成羰基化合物$\rm Ni(CO)_4$；为氧化还原反应](1)</Checkpoint>

<Checkpoint>[剩余一分子$\rm CO$充当还原剂](1)</Checkpoint>

$\rm NiS + 5CO + 4NaOH = Ni(CO)_4 + Na_2CO_3+ Na_2S + 2H_2O$

<Checkpoint>[$\rm NiS + 5CO + 4NaOH = Ni(CO)_4 + Na_2CO_3+ Na_2S + 2H_2O$](1)</Checkpoint>

反应物系数和为10，产物系数和为5，商为整数2，选项F错误。

选项B-F均错误，选项A正确。