1. Under anaerobic conditions, a $\mathrm { C u } ^ { 2 + }$ containing solution can act as an oxidizing agent to react with pyrite, and the product contains sulfur in two oxidation states: -2 and +6; the oxidation state of iron remains unchanged.
2. $\mathrm { L i F e P O _ { 4 } }$ is prepared by mixing and grinding iron phosphate, lithium carbonate, and oxalic acid dihydrate, followed by high-temperature calcination.
3. Under weakly acidic conditions, reacting $\mathrm { L i F e P O _ { 4 } }$ with hydrogen peroxide aqueous solution can make all the lithium enter the aqueous solution, while iron and phosphorus remain in the solid phase.
4. Elemental mercury, alum, and $\mathrm{NaCl}$ are mixed and heated in air to volatilize a compound containing mercury in the +1 oxidation state.
5. In acidic aqueous solution, potassium hexathiocyanatochromate(III) is oxidized with excess $\mathrm { I } _ { 2 } $, and carbon and nitrogen are only converted to $\mathrm{ICN}$; no element is produced in the reaction.

Write out the equations of the above 5 reactions (1, 3, and 5 are ionic equations), requiring the coefficients of each substance to be the simplest integer ratio. Which of the following options is correct?
1，在无氧条件下，含 $\mathrm { C u } ^ { 2 + }$ 的溶液可作为氧化剂与黄铁矿反应，产物中硫有2种氧化态：-2和 + 6 ；铁的氧化态保持不变。  
2，将磷酸铁、碳酸锂以及二水合草酸混合研磨后高温煅烧制备 $\mathrm { L i F e P O _ { 4 } }$ 。  
3，在弱酸性条件下，用过氧化氢水溶液与 $\mathrm { L i F e P O _ { 4 } }$ 反应，能使得锂全部进入水溶液中，而铁和磷仍留在固相。  
4，将汞单质、明矾和$\mathrm{NaCl}$在空气中混合加热，挥发出含+1氧化态汞的化合物。  
5，在酸性水溶液中，用过量的 $\mathrm { I } _ { 2 }$ 氧化六硫氰酸根合铬(III)酸钾，碳和氮只转化为$\mathrm{ICN}$；反应无单质产生。

书写出以上5个反应的方程式（1, 3和5为离子方程式），要求各物质系数为最简整数比，下列选项正确的是：
All other options are incorrect
The ionic equation for reaction 1, where the sum of the product coefficients is less than the sum of the reactant coefficients.
The chemical equation for reaction 2, where the sum of the product coefficients is not a multiple of the sum of the reactant coefficients.
The ionic equation for reaction 3, the sum of coefficients of all species is a composite number.
The chemical equation for reaction 4, where the sum of the coefficients of all species is greater than 100.
The ionic equation for reaction 5, the sum of the coefficients of all species is greater than 150.
其他选项均不正确
反应1的离子方程式，产物系数和小于反应物系数和
反应2的化学方程式，产物系数和不为反应物系数和的倍数
反应3的离子方程式，所有物种系数和为合数
反应4的化学方程式，所有物种系数和大于100
反应5的离子方程式，所有物种的系数和大于150
Reaction 1, the iron in pyrite is +2 valence, the unchanged oxidation state indicates the generation of $\mathrm{Fe^{2+}}$, the +6 valence sulfur should be sulfate, and the -2 valence sulfur should form $\mathrm{Cu_2S}$ with the reduced copper ions. Thus, the balanced equation can be written as:

$\mathrm{14Cu^{2+}+5FeS_2+12H_2O\to 3SO_4^{2-}+7Cu_2S+5Fe^{2+}+24H^+}$

<Checkpoint>[$\mathrm{14Cu^{2+}+5FeS_2+12H_2O\to 3SO_4^{2-}+7Cu_2S+5Fe^{2+}+24H^+}$](1)</Checkpoint>

The sum of product coefficients is 39, and the sum of reactant coefficients is 31, so option B is incorrect.

Reaction 2, the products other than lithium iron phosphate should be carbon dioxide and water. Thus, the balanced equation can be written as:

$\mathrm{Li_2CO_3+2FePO_4+H_2C_2O_4\cdot 2H_2O\to 2LiFePO_4+3H_2O+3CO_2}$

<Checkpoint>[$\mathrm{Li_2CO_3+2FePO_4+H_2C_2O_4\cdot 2H_2O\to 2LiFePO_4+3H_2O+3CO_2}$](1)</Checkpoint>

The sum of product coefficients is 8, and the sum of reactant coefficients is 4, so option C is incorrect.

Reaction 3, hydrogen peroxide acts as an oxidant to oxidize ferrous, thereby allowing lithium to enter the solution. The remaining products are iron phosphate and water. Thus, the balanced equation can be written as:

<Checkpoint>[Hydrogen peroxide acts as an oxidant to oxidize ferrous](1)</Checkpoint>

$\mathrm{2LiFePO_4+2H^++H_2O_2\to 2FePO_4+2Li^++2H_2O}$

<Checkpoint>[$\mathrm{2LiFePO_4+2H^++H_2O_2\to 2FePO_4+2Li^++2H_2O}$](1)</Checkpoint>

The sum of coefficients of all species is 11, which is a prime number, not a composite number, so option D is incorrect.

Reaction 4, alum is $\mathrm{KAl(SO_4)_2\cdot 12H_2O}$, and the gaseous compound of +1 valence mercury should be $\mathrm{Hg_2Cl_2}$. Aluminum turns into aluminum oxide, and potassium and sodium turn into sulfates. Thus, the balanced equation can be written as:

<Checkpoint>[The gaseous compound of +1 valence mercury should be $\mathrm{Hg_2Cl_2}$](1)</Checkpoint>

$\mathrm{12Hg+3O_2+12NaCl+4[KAl(SO_4)_2\cdot 12H_2O]\to 6Hg_2Cl_2+2Al_2O_3+2K_2SO_4+6Na_2SO_4+48H_2O}$

<Checkpoint>[$\mathrm{12Hg+3O_2+12NaCl+4[KAl(SO_4)_2\cdot 12H_2O]\to 6Hg_2Cl_2+2Al_2O_3+2K_2SO_4+6Na_2SO_4+48H_2O}$](1)</Checkpoint>

The sum of coefficients of all species is 95, so option E is incorrect.

Reaction 5, elemental iodine cannot oxidize +3 valence chromium to +6 valence, so the only thing that can be oxidized is sulfur, and the product is sulfate. The remaining products are given in the question. Thus, the balanced equation can be written as:

<Checkpoint>[Elemental iodine cannot oxidize +3 valence chromium to +6 valence, the only thing that can be oxidized is sulfur](1)</Checkpoint>

$\mathrm{[Cr(SCN)_6]^{3-}+24I_2+24H_2O\to Cr^{3+}+6SO_4^{2-}+6ICN+48H^++42I^-}$

<Checkpoint>[$\mathrm{[Cr(SCN)_6]^{3-}+24I_2+24H_2O\to Cr^{3+}+6SO_4^{2-}+6ICN+48H^++42I^-}$](1)</Checkpoint>

The sum of coefficients of all species is 152, so option F is correct.
反应1，黄铁矿中铁为+2价，氧化态不变表明生成了$\mathrm{Fe^{2+}}$，+6价硫应该为硫酸根，-2价硫应该与被还原的铜离子组成$\mathrm{Cu_2S}$，由此便可写出配平后的方程式：

$\mathrm{14Cu^{2+}+5FeS_2+12H_2O\to 3SO_4^{2-}+7Cu_2S+5Fe^{2+}+24H^+}$

<Checkpoint>[$\mathrm{14Cu^{2+}+5FeS_2+12H_2O\to 3SO_4^{2-}+7Cu_2S+5Fe^{2+}+24H^+}$](1)</Checkpoint>

产物系数和为39，反应物系数和为31，选项B错误。

反应2磷酸亚铁锂之外的产物应该为二氧化碳和水，由此可写出配平后的方程式：

$\mathrm{Li_2CO_3+2FePO_4+H_2C_2O_4\cdot 2H_2O\to 2LiFePO_4+3H_2O+3CO_2}$

<Checkpoint>[$\mathrm{Li_2CO_3+2FePO_4+H_2C_2O_4\cdot 2H_2O\to 2LiFePO_4+3H_2O+3CO_2}$](1)</Checkpoint>

产物系数和为8，反应物系数和为4，选项C错误。

反应3过氧化氢作为氧化剂将亚铁氧化从而使锂进入溶液，剩下的产物为磷酸铁和水，由此可以写出配平后的方程式：

<Checkpoint>[过氧化氢作为氧化剂将亚铁氧化](1)</Checkpoint>

$\mathrm{2LiFePO_4+2H^++H_2O_2\to 2FePO_4+2Li^++2H_2O}$

<Checkpoint>[$\mathrm{2LiFePO_4+2H^++H_2O_2\to 2FePO_4+2Li^++2H_2O}$](1)</Checkpoint>

所有物种系数和为11，是质数不是合数，选项D错误。

反应4，明矾为$\mathrm{KAl(SO_4)_2\cdot 12H_2O}$，+1价汞的气态化合物应该为$\mathrm{Hg_2Cl_2}$，铝变为氧化铝，钾和钠变为硫酸盐，由此可写出配平后的方程式：

<Checkpoint>[+1价汞的气态化合物应该为$\mathrm{Hg_2Cl_2}$](1)</Checkpoint>

$\mathrm{12Hg+3O_2+12NaCl+4[KAl(SO_4)_2\cdot 12H_2O]\to 6Hg_2Cl_2+2Al_2O_3+2K_2SO_4+6Na_2SO_4+48H_2O}$

<Checkpoint>[$\mathrm{12Hg+3O_2+12NaCl+4[KAl(SO_4)_2\cdot 12H_2O]\to 6Hg_2Cl_2+2Al_2O_3+2K_2SO_4+6Na_2SO_4+48H_2O}$](1)</Checkpoint>

所有物种系数和为95，选项E错误。

反应5，碘单质无法将+3价的铬氧化至+6价，因此被氧化的只可能是硫，产物为硫酸根，剩下的产物题目中已经给出，由此可以写出配平后的方程式：

<Checkpoint>[碘单质无法将+3价的铬氧化至+6价，被氧化的只可能是硫](1)</Checkpoint>

$\mathrm{[Cr(SCN)_6]^{3-}+24I_2+24H_2O\to Cr^{3+}+6SO_4^{2-}+6ICN+48H^++42I^-}$

<Checkpoint>[$\mathrm{[Cr(SCN)_6]^{3-}+24I_2+24H_2O\to Cr^{3+}+6SO_4^{2-}+6ICN+48H^++42I^-}$](1)</Checkpoint>

所有物种系数和为152，选项F正确。