The wastewater produced during copper smelting contains a high concentration of arsenic, mainly in the form of $AsO_2^-$. The following treatments were all carried out at room temperature.

First, add a $26\%$ $Na_2S$ solution, adjust the $pH$ to $0.8$ with sulfuric acid, and after reacting for twenty minutes, a precipitate is obtained, which is recorded as reaction 1.

Subsequently, the precipitate is added to a slightly excessive amount of $NaOH$ solution at a molar ratio of $1:7$ to ensure complete dissolution of the precipitate, which is recorded as reaction 2.

After dissolution, a small amount of hydroquinone is added as a catalyst and sodium dodecyl sulfonate as a surfactant, and air is passed through for ten hours for oxidation and desulfurization, which is recorded as reaction 3.

After oxidation, the $pH$ is adjusted to $0$ with sulfuric acid, and $SO_2$ is then passed through to generate a precipitate, which is recorded as reaction 4.

Finally, the precipitate is filtered to obtain a solid oxide M with high purity. Detection reveals that the final remaining filtrate still contains a certain amount of $Re$ element that can be recovered.

The above filtrate is extracted with an organic phase composed of $20\%$ trioctylamine, $40\%$ sec-octyl alcohol, and $40\%$ kerosene. After liquid separation, the organic phase is taken and extracted with ammonia water. After liquid separation again, the aqueous phase is frozen, crystallized, and separated to obtain a pure compound $A$ without water of crystallization.

$1.3874g$ of compound $A$ is taken and gradually heated to $800 \degree \mathrm{C}$ in a $H_2$ atmosphere, and the final solid product $B$ is $0.9631g\;Re$, and $0.3727g$ of water vapor is separated in the tail gas. If heated in an air atmosphere, $100\%$ weight loss is observed at $400\degree \mathrm{C}$.

Regarding the above process, the following descriptions are given. Let the product of the serial numbers of all incorrect descriptions be $x$, and the sum of the serial numbers of all correct descriptions be $y$. Find $\dfrac{x}{(y+10)^2}$ and select the option with an absolute error $<0.001$. If there is no option that meets the condition, select $A$.

1.  It is not possible to select four coefficients from the equation for reaction 1 that form an arithmetic sequence.

2.  The sum of the coefficients on the left side of the equation for reaction 2 divided by the sum of the coefficients on the right side is equal to 1.4.

3.  The average coefficient of the equation for reaction 3 is equal to 3.8.

4.  The product of all the coefficients on both sides of the equation for reaction 4 is equal to 144.

5.  The catalytic principle of hydroquinone is the generation of a p-benzoquinone intermediate.

6.  One molecule of compound $A$ contains 10 atoms.

7.  One molecule of compound $A$ contains 9 atoms.

8.  When compound $A$ is treated at $800\degree \mathrm{C}$ in a $H_2$ atmosphere, one molecule of compound $A$ generates 4 molecules of gas.

9.  When treated at $400\degree \mathrm{C}$ in an air atmosphere, a reaction with oxygen occurs.

10. Organic phase extraction utilizes sec-octyl alcohol and kerosene to suppress the formation of a third phase.

11. Back extraction cannot use a high $pH$ solution.
铜冶炼过程所产生的废水中含有较高浓度的砷元素，主要以 $AsO_2^-$ 形式存在。以下处理均在室温下进行。

首先加入 $26\%$的 $Na_2S$ 溶液，用硫酸调节 $pH=0.8$，反应二十分钟后得到沉淀，记为反应1。

随后按物质的量之比 $1:7$ 将沉淀加入到略过量的 $NaOH$ 溶液中，可保证沉淀完全溶解，记为反应2。

溶解后加入少量的对苯二酚作为催化剂和十二烷基磺酸钠作为表面活性剂，通空气十小时氧化脱硫，记为反应3。

氧化后，用硫酸调节 $pH=0$，再通 $SO_2$ 生成沉淀，记为反应4。

最后过滤得到沉淀为纯度较高的固体氧化物M。检测发现最终剩余的滤液中还含有一定量的 $Re$ 元素可以回收。

用 $20 \%$ 的三辛基胺、 $4 0 \%$ 仲辛醇和 $4 0 \%$ 煤油组成的有机相萃取上述滤液，分液后取有机相用氨水萃取，再次分液后将水相冷冻结晶分离得到纯净的不含结晶水的化合物 $A$。

取 $1.3874g$ 化合物 $A$ 在 $H_2$ 氛围下逐渐加热至 $800 \degree \mathrm{C}$ ，最终得到固体产物 $B$ 为 $0.9631g\;Re$，并在尾气中分离到 $0.3727g$ 水蒸气。如果在空气氛围中加热，至 $400\degree \mathrm{C}$ 时观察到失重 $1 0 0 \%$ 。

针对上述过程给出如下叙述，记所有错误叙述序号的乘积为 $x$，所有正确叙述序号的和为 $y$，求 $\dfrac{x}{(y+10)^2}$，选出误差绝对值 $<0.001$ 的选项，如果没有符合条件的选项则选择选项 $A$。

1.反应1的方程式两侧系数中任意4个不能构成等差数列

2.反应2的方程式左侧系数之和除以右侧系数之和等于1.4

3.反应3的方程式两侧所有系数平均数为 3.8

4.反应4的方程式两侧所有系数相乘等于144

5.对苯二酚催化原理是生成了对苯醌中间体

6.一分子化合物 $A$ 含10个原子

7.一分子化合物 $A$ 含9个原子

8.$H_2$ 氛围下 $800\degree \mathrm{C}$ 处理时一分子化合物 $A$ 生成4分子气体

9.空气氛围下 $400\degree \mathrm{C}$ 处理时与氧气发生了反应

10.有机相萃取利用仲辛醇和煤油抑制第三相的形成

11.反萃取不能使用高 $pH$ 溶液
All other options are incorrect
1.429
0.802
1.565
0.913
1.905
0.634
2.665
其他选项均不正确
1.429
0.802
1.565
0.913
1.905
0.634
2.665
The equations for reactions 1 to 4 are as follows:

$2AsO_2^- + 3S^{2-} + 8H^+ \rightarrow As_2S_3+ 4H_2O$ 

$As_2S_3 + 6OH^- \rightarrow AsO_3^{3-} + AsS_3^{3-} + 3H_2O$ 

$2AsO_3^{3-} + 2AsS_3^{3-} + 5O_2 \rightarrow  4AsO_4^{3-} + 6S$ 

$2AsO_4^{3-} + 2SO_2+ 2H^+ \rightarrow  As_2O_3+ 2SO_4^{2-} + H_2O$ 

The coefficients $1, 2, 3, 4$ in the equation of reaction 1 constitute an arithmetic sequence with a common difference of 1, so statement 1 is incorrect.

<Checkpoint>[Equation for reaction 1 $2AsO_2^- + 3S^{2-} + 8H^+ \rightarrow As_2S_3+ 4H_2O$](0.5)</Checkpoint>

The sum of the coefficients on the left side of the equation for reaction 2, 7, divided by the sum of the coefficients on the right side, 5, equals 1.4, so statement 2 is correct.

<Checkpoint>[Equation for reaction 2 $As_2S_3 + 6OH^- \rightarrow AsO_3^{3-} + AsS_3^{3-} + 3H_2O$](0.5)</Checkpoint>

The average coefficient of the equation for reaction 3 is $\dfrac{2+2+5+4+6}{5}=3.8$, so statement 3 is correct.

<Checkpoint>[Equation for reaction 3 $2AsO_3^{3-} + 2AsS_3^{3-} + 5O_2 \rightarrow  4AsO_4^{3-} + 6S$](0.5)</Checkpoint>

The product of all the coefficients on both sides of the equation for reaction 4 is equal to 16, so statement 4 is incorrect.

<Checkpoint>[Equation for reaction 4 $2AsO_4^{3-} + 2SO_2+ 2H^+ \rightarrow  As_2O_3+ 2SO_4^{2-}+ H_2O$](0.5)</Checkpoint>

The first step of hydroquinone catalysis is oxidation to benzoquinone by oxygen, so statement 5 is correct.

<Checkpoint>[In the first step of catalysis, hydroquinone is oxidized to benzoquinone](0.5)</Checkpoint>

The molar mass of compound $A$ is $M_A =\dfrac{m_AAr(Re)}{m_B}=\dfrac{1.3874\times186.2}{0.9631}=268.2 \;g/mol$ 

The molar ratio of $Re$ and $O$ in compound $A$ is $n(Re):n(O)=\dfrac{m_B}{Ar(Re)}:\dfrac{m_{H_2O}}{Mr(H_2O)}=\dfrac{0.9631}{186.2}:\dfrac{0.3727}{18.02}=1:3.999\approx1:4$ 

The remaining element $Mr=268.2-186.2-16.00\times4=18.0$, which is inferred to be $NH_4^+$

Therefore, the chemical formula of compound $Α$ is $NH_4ReO_4$, which contains 10 atoms in one molecule, so statement 6 is correct and statement 7 is incorrect.

<Checkpoint>[Compound $Α$ is $NH_4ReO_4$](1)</Checkpoint>

The equation for treatment at $800\degree \mathrm{C}$ under $H_2$ atmosphere is $2NH_4ReO_4 + 7H _2 \rightarrow 2NH_3 + 2Re + 8H_2O$, so one molecule of $A$ generates 5 molecules of gas, so statement 8 is incorrect.

<Checkpoint>[Hydrogen atmosphere equation $2NH_4ReO_4 + 7H _2 \rightarrow 2NH_3 + 2Re + 8H_2O$](0.5)</Checkpoint>

The equation for treatment at $400\degree \mathrm{C}$ under air atmosphere is $2NH_4ReO_4 \rightarrow 2NH_3 + Re_2O_7 + H_2O$, which is the self-decomposition of compound $A$, without the participation of oxygen, so statement 9 is incorrect.

<Checkpoint>[Air atmosphere equation $2NH_4ReO_4 \rightarrow 2NH_3 + Re_2O_7 + H_2O$](0.5)</Checkpoint>

The equation for organic phase extraction is $R_3N\text{(organic phase)} + ReO_4^-\text{(aqueous phase)} + H^+\text{(aqueous phase)} \rightarrow R_3NH\cdot ReO_4\text{(organic phase)}$, in the organic phase, trioctylamine is used to affinity the required product, and sec-octyl alcohol and kerosene are used to enhance the uniformity of the organic phase and avoid the precipitation of the formed ion pairs into a third phase, so statement 10 is correct.

<Checkpoint>[sec-octyl alcohol and kerosene avoid the precipitation of the formed ion pairs into a third phase](0.5)</Checkpoint>

The equation for ammonia back-extraction is $R_3NH·ReO_4\text{(organic phase)} + NH_3\text{(aqueous phase)} \rightarrow R_3N\text{(organic phase)} + NH_4^+\text{(aqueous phase)} + ReO_4^-\text{(aqueous phase)}$, if a high $pH$ solution is used, other cations may be introduced or the extraction reaction may be reversed, reducing the recovery rate, so statement 11 is correct.

<Checkpoint>[During back-extraction, a high $pH$ solution may introduce other cations or reverse the extraction reaction](0.5)</Checkpoint>

All correct statements are $2,3,5,6,10,11$, calculate $y=2+3+5+6+10+11=37$; all incorrect statements are $1,4,7,8,9$, calculate $x=1\times4\times7\times8\times9=2016$. Then $\dfrac{x}{(y+10)^2}=\dfrac{2016}{(37+10)^2}=0.91263\approx0.913$, so option $E$ is correct.

<Checkpoint>[$x=2016,y=37,\dfrac{x}{(y+10)^2}=0.913$](0.5)</Checkpoint>
反应1到4的方程式如下：

$2AsO_2^- + 3S^{2-} + 8H^+ \rightarrow As_2S_3+ 4H_2O$ 

$As_2S_3 + 6OH^- \rightarrow AsO_3^{3-} + AsS_3^{3-} + 3H_2O$ 

$2AsO_3^{3-} + 2AsS_3^{3-} + 5O_2 \rightarrow  4AsO_4^{3-} + 6S$ 

$2AsO_4^{3-} + 2SO_2+ 2H^+ \rightarrow  As_2O_3+ 2SO_4^{2-} + H_2O$ 

反应1的方程式中系数 $1,2,3,4$ 构成公差为1的等差数列，叙述1错误。

<Checkpoint>[反应1方程式 $2AsO_2^- + 3S^{2-} + 8H^+ \rightarrow As_2S_3+ 4H_2O$](0.5)</Checkpoint>

反应2的方程式左侧系数之和7除以右侧系数之和5等于1.4，叙述2正确。

<Checkpoint>[反应2方程式 $As_2S_3 + 6OH^- \rightarrow AsO_3^{3-} + AsS_3^{3-} + 3H_2O$](0.5)</Checkpoint>

反应3的方程式系数平均数 $\dfrac{2+2+5+4+6}{5}=3.8$，叙述3正确。

<Checkpoint>[反应3方程式 $2AsO_3^{3-} + 2AsS_3^{3-} + 5O_2 \rightarrow  4AsO_4^{3-} + 6S$](0.5)</Checkpoint>

反应4的方程式两侧系数依次相乘等于16，叙述4错误。

<Checkpoint>[反应4方程式 $2AsO_4^{3-} + 2SO_2+ 2H^+ \rightarrow  As_2O_3+ 2SO_4^{2-} + H_2O$](0.5)</Checkpoint>

对苯二酚催化第一步为被氧气氧化为对苯醌，叙述5正确。

<Checkpoint>[催化第一步对苯二酚被氧化为对苯醌](0.5)</Checkpoint>

化合物 $A$ 的摩尔质量 $M_A =\dfrac{m_AAr(Re)}{m_B}=\dfrac{1.3874\times186.2}{0.9631}=268.2 \;g/mol$ 

化合物 $A$ 中 $Re$ 和 $O$ 的摩尔比  $n(Re):n(O)=\dfrac{m_B}{Ar(Re)}:\dfrac{m_{H_2O}}{Mr(H_2O)}=\dfrac{0.9631}{186.2}:\dfrac{0.3727}{18.02}=1:3.999\approx1:4$ 

其余元素$Mr=268.2-186.2-16.00\times4=18.0$ ，推测为$NH_4^+$

则化合物 $Α$ 的化学式为 $NH_4ReO_4$，一分子中含10个原子，叙述6正确而叙述7错误。

<Checkpoint>[化合物 $Α$ 为 $NH_4ReO_4$](1)</Checkpoint>

 $H_2$ 氛围下 $800\degree \mathrm{C}$ 处理方程式为 $2NH_4ReO_4 + 7H _2 \rightarrow 2NH_3 + 2Re + 8H_2O$，所以一分子 $A$ 生成5分子气体，叙述8错误。

<Checkpoint>[氢气氛围方程式 $2NH_4ReO_4 + 7H _2 \rightarrow 2NH_3 + 2Re + 8H_2O$](0.5)</Checkpoint>

空气氛围下 $400\degree \mathrm{C}$ 处理时方程式为 $2NH_4ReO_4 \rightarrow 2NH_3 + Re_2O_7 + H_2O$，是化合物 $A$ 自身分解，没有氧气的参与，叙述9错误。

<Checkpoint>[空气氛围方程式 $2NH_4ReO_4 \rightarrow 2NH_3 + Re_2O_7 + H_2O$](0.5)</Checkpoint>

有机相萃取的方程式为 $R_3N\text{(有机相)} + ReO_4^-\text{(水相)} + H^+\text{(水相)} \rightarrow R_3NH\cdot ReO_4\text{(有机相)}$，有机相中三辛基胺用于亲和需要的产物，仲辛醇和煤油用于增强有机相的均一性，避免形成的离子对析出成为第三相，叙述10正确。

<Checkpoint>[仲辛醇和煤油避免形成的离子对析出成为第三相](0.5)</Checkpoint>

氨水反萃取的方程式为 $R_3NH·ReO_4\text{(有机相)} + NH_3\text{(水相)} \rightarrow R_3N\text{(有机相)} + NH_4^+\text{(水相)} + ReO_4^-\text{(水相)}$，如果使用高 $pH$ 溶液，可能引入其他阳离子或逆转萃取反应，降低回收率，叙述11正确。

<Checkpoint>[反萃取时高 $pH$ 溶液可能引入其他阳离子或逆转萃取反应](0.5)</Checkpoint>

所有正确叙述有 $2,3,5,6,10,11$，计算 $y=2+3+5+6+10+11=37$；所有错误叙述有 $1,4,7,8,9$，计算 $x=1\times4\times7\times8\times9=2016$。则 $\dfrac{x}{(y+10)^2}=\dfrac{2016}{(37+10)^2}=0.91263\approx0.913$，选项 $E$ 正确。

<Checkpoint>[$x=2016,y=37,\dfrac{x}{(y+10)^2}=0.913$](0.5)</Checkpoint>