The FeSe-based system is the simplest type of iron-based superconducting material. The filling of iron vacancies with $\mathrm{Cu}$ in the form of a univalent cation will affect the structure and electrical and magnetic properties of the superconductor. In this experiment, precipitation gravimetry and $\mathrm{EDTA}$ complexometric titration were used to determine the chemical composition of a $\mathrm{Fe-Cu-Se}$ superconductor. The specific experimental procedure is as follows:

(a) Accurately weigh approximately $0.4639~\mathrm{g}$ of the sample into a $100~\mathrm{mL}$ beaker, add $6~\mathrm{mL}$ of concentrated nitric acid and $6~\mathrm{mL}$ of concentrated hydrochloric acid, and heat. Add concentrated hydrochloric acid during heating until the sample is completely dissolved. Add $20~\%$ hydroxylamine hydrochloride solution dropwise under stirring and heating until precipitation is complete, keep warm for a period of time, carefully filter out the precipitate with pre-weighed filter paper, wash several times with deionized water, dry at low temperature until constant weight, and record the mass as $0.2565~\mathrm{g}$.

(b) Weigh approximately $0.8527~\mathrm{g}$ of the sample into a $100~\mathrm{mL}$ beaker, add $10~\mathrm{mL}$ of concentrated nitric acid and $10~\mathrm{mL}$ of concentrated hydrochloric acid, and heat. Add concentrated hydrochloric acid during heating until the sample is completely dissolved. Concentrate the solution to about $10~\mathrm{mL}$, and dilute to $100~\mathrm{mL}$ with deionized water. Transfer $25.00~\mathrm{mL}$ of the above solution to a conical flask, and adjust the $\mathrm{pH}$ to $3\sim4$ with $1:1$ ammonia water; add $5~\mathrm{g}$ of solid ammonium fluoride, shake to dissolve, then add $5~\mathrm{mL}$ of $2~\mathrm{mol/L}$ acetic acid-sodium acetate buffer solution, $5$ drops of $\mathrm{PAN}$ indicator, and titrate slowly with $0.02035~\mathrm{mol/L}~\mathrm{EDTA}$ until the endpoint is reached when the purple-red color changes to gray-blue-green, consuming $13.94 ~\mathrm{mL}$.

(c) Transfer $10.00~\mathrm{mL}$ of the $100~\mathrm{mL}$ solution prepared in (b) to a $100~\mathrm{mL}$ volumetric flask and dilute to the mark with distilled water. Transfer $25.00~\mathrm{mL}$ of the solution to a small beaker, add $7~\mathrm{mL}$ of $1:1$ ammonia water while stirring, heat for a period of time, and filter under normal pressure after the precipitate is obviously agglomerated. Wash the precipitate three times with deionized water. Discard the filtrate. Dissolve the precipitate with $15~\mathrm{mL}$ of $0.4\mathrm{mol/L}$ sulfuric acid, collect the solution in a conical flask, add $10~\mathrm{mL}$ of deionized water, $5$ drops of sulfosalicylic acid indicator, and titrate with $0.008140~\mathrm{mol/L}~\mathrm{EDTA}$ until the purple-red color fades significantly, which is the endpoint, consuming $14.53~\mathrm{mL}$.

For this experiment, the following options are correct (for options involving calculations, consider errors within $0.1~\%$ as correct)
以 $\mathrm{FeSe}$ 为代表的体系是最简单的一类铁基超导材料，$\mathrm{Cu}$ 以正一价形式填入铁空位中， 将对超导体的结构和电学、磁学性质都会产生影响。本实验利用沉淀重量法、$\mathrm{EDTA}$ 络合滴定法测定了一种 $\mathrm{Fe-Cu-Se}$ 超导体的化学组成。具体的实验过程如下：

(a) 准确称取约 $0.4639~\mathrm{g}$ 样品于 $100~\mathrm{mL}$ 烧杯中，加入 $6~\mathrm{mL}$ 浓硝酸、$6~\mathrm{mL}$ 浓盐酸后加热。加热过程中补加浓盐酸至样品全部溶解。搅拌加热下滴加 $20~\%$ 盐酸羟胺溶液至沉淀完全，保温一段时间，用已恒重的滤纸小心滤出沉淀，用去离子水洗涤几次，低温下烘干至恒重，记录质量 $0.2565~\mathrm{g}$。

(b) 称取约 $0.8527~\mathrm{g}$ 样品于 $100~\mathrm{mL}$ 烧杯中，加入 $10~\mathrm{mL}$ 浓硝酸、$10~\mathrm{mL}$ 浓盐酸后加热。加热过程中补加浓盐酸至样品全部溶解。将溶液浓缩至 $10~\mathrm{mL}$ 左右，用去离子水定容至 $100~\mathrm{mL}$。移取 $25.00~\mathrm{mL}$ 上述溶液于锥形瓶中，用$1:1$ 的氨水调节 $\mathrm{pH}=3\sim4$；加入 $5~\mathrm{g}$ 固体氟化铵，摇动使之溶解后加入 $5~\mathrm{mL}~2~\mathrm{mol/L}$ 醋酸-醋酸钠缓冲溶液，$5$ 滴 $\mathrm{PAN}$ 指示剂，用 $0.02035~\mathrm{mol/L}~\mathrm{EDTA}$ 慢速滴定至紫红色经灰色变为蓝绿色即为终点，消耗 $13.94 ~\mathrm{mL}$。

(c) 移取(b)中定容好的 $100~\mathrm{mL}$ 溶液中的 $10.00~\mathrm{mL}$ 于 $100~\mathrm{mL}$ 容量瓶中，用蒸馏水定容。移取 $25.00~\mathrm{mL}$ 溶液于小烧杯中，搅拌下加入 $7~\mathrm{mL}~1:1$ 氨水，加热一段时间，待沉淀明显团聚后常压过滤。用去离子水洗涤沉淀三次。弃去滤液。沉淀用 $15~\mathrm{mL}~0.4\mathrm{mol/L}$ 硫酸溶解，溶液收集在锥形瓶中，加入 $10~\mathrm{mL}$ 去离子水，$5$ 滴磺基水杨酸指示剂，用$0.008140~\mathrm{mol/L}~\mathrm{EDTA}$ 滴定至紫红色明显褪去即为终点，消耗 $14.53~\mathrm{mL}$ 。

对于这个实验，下列选项正确的是（涉及到计算的选项，视误差 $0.1~\%$ 之内为正确）
Ascorbic acid thiourea can be used to mask $\mathrm{Cu}$ to measure $\mathrm{Fe}$
Cyanide masking of $\mathrm{Cu}$ can be used to measure $\mathrm{Fe}$.
Sulfur ion masking of $\mathrm{Cu}$ can be used to measure $\mathrm{Fe}$.
Hydrochloric acid + hydrogen peroxide can be used as a complete replacement for nitric acid + hydrochloric acid.
(a) In experiments, high-temperature drying can be used to speed up the precipitation process.
(a) If there is no holding period at a constant temperature during the experiment, the calculated mass fraction of $\mathrm{Se}$ will be higher.
(c) The purpose of adding ammonia in the experiment is to adjust the $\mathrm{pH}$ to slightly acidic.
(c) The titration temperature in the experiment was room temperature.
Representing the alloy with $\mathrm{Fe}_x\mathrm{Cu}_y\mathrm{Se}$, then $x+y=0.8969$
The ratio of ferrous iron to ferric iron is approximately $2.516:1$.
The mass fraction of $\mathrm{Se}$ is 1.954 times that of $\mathrm{Fe}$.
All of the above options are incorrect.
可以采用抗坏血酸硫脲掩蔽 $\mathrm{Cu}$ 以测量 $\mathrm{Fe}$ 
可以采用氰离子掩蔽 $\mathrm{Cu}$ 以测量 $\mathrm{Fe}$ 
可以采用硫离子掩蔽 $\mathrm{Cu}$ 以测量 $\mathrm{Fe}$
可以采用盐酸+过氧化氢完全替代硝酸+盐酸
(a)实验中为了加快速度可以高温烘干沉淀
(a)实验中如果没有保温一段时间，计算出的 $\mathrm{Se}$ 质量分数将变高
(c)实验中加入氨水的目的是调整 $\mathrm{pH}$ 为微酸性
(c)实验中滴定温度为室温
用 $\mathrm{Fe}_x\mathrm{Cu}_y\mathrm{Se}$ 表示该合金，则 $x+y=0.8969$
二价铁与三价铁比例约 $2.516:1$ 
$\mathrm{Se}$ 的质量分数是 $\mathrm{Fe}$ 的 $1.954$倍
以上选项均不正确
Ascorbic acid thiourea, cyanide ions, and sulfide ions masking $\mathrm{Cu}$ will all react with $\mathrm{Fe}^{3+}$, so they cannot be used, therefore options A, B, and C are all incorrect.

<Checkpoint>[Because ascorbic acid thiourea, cyanide ions, and sulfide ions all react with iron, options A, B, and C are all incorrect](1)</Checkpoint>

Hydrogen peroxide decomposes upon heating, and $\mathrm{Fe}^{3+}$ catalyzes the decomposition of hydrogen peroxide, consuming too much hydrogen peroxide; and a small amount of nitric acid must be added to completely dissolve the sample, so it cannot be replaced, option D is incorrect.

<Checkpoint>[Because hydrogen peroxide is catalyzed by iron and decomposes upon heating, and nitric acid needs to be added to dissolve the sample, option D is incorrect](1)</Checkpoint>

(a) In the experiment, high-temperature drying of the precipitate will cause Se to sublime, resulting in less precipitate obtained, option E is incorrect.

<Checkpoint>[High-temperature drying of the precipitate causes selenium to sublime, losing precipitate, option E is incorrect](1)</Checkpoint>

(a) In the experiment, if it is not kept warm for a period of time, the precipitate will not be sufficiently aged, resulting in less precipitate obtained, and the calculated mass fraction of $\mathrm{Se}$ will be lower, option F is incorrect.

<Checkpoint>[Incomplete aging of the precipitate will result in loss of precipitate, option F is incorrect](0.5)</Checkpoint>

<Checkpoint>[The calculated mass fraction of $\mathrm{Se}$ will be lower, option F is incorrect](0.5)</Checkpoint>

(c) In the experiment, ammonia is added to adjust the $\mathrm{pH}$ to slightly alkaline to completely precipitate $\mathrm{Fe}$. A slightly acidic environment will result in less precipitate obtained, option G is incorrect.

<Checkpoint>[Iron needs to precipitate in a slightly alkaline environment, option G is incorrect](0.5)</Checkpoint>

(c) In the experiment, the titration temperature is $60-70$ degrees Celsius, because the reaction rate of $\mathrm{EDTA}$ and $\mathrm{Fe}$ is very slow at room temperature, causing the indicator to become sluggish. Therefore, the temperature is raised to increase the reaction rate of $\mathrm{EDTA}$ and $\mathrm{Fe}$ and avoid the indicator becoming sluggish, option H is incorrect.

<Checkpoint>[The reaction rate of $\mathrm{EDTA}$ and $\mathrm{Fe}$ is very slow at room temperature, causing the indicator to become sluggish, option H is incorrect](1)</Checkpoint>

(a) Hydroxylamine hydrochloride precipitates $\mathrm{Se}$, so the content of $\mathrm{Se}$ is determined by gravimetric analysis.

<Checkpoint>[(a) Hydroxylamine hydrochloride precipitates $\mathrm{Se}$](1)</Checkpoint>

The weight of selenium in the alloy sample is $m_{\mathrm{Se}}=0.2565~\mathrm{g}$

The mass fraction of selenium in the alloy sample is $\omega_{\mathrm{Se}}=\dfrac{0.2565~\mathrm{g}}{0.4639~\mathrm{g}}=55.2921~\%$

<Checkpoint>[$\omega_{Se}=55.29~\%$ ](0.5)</Checkpoint>

(b) Ammonium fluoride masks $\mathrm{Fe}^{3+}$, so the content of $\mathrm{Cu}$ is obtained by $\mathrm{EDTA}$ titration.

<Checkpoint>[(b) Ammonium fluoride masks $\mathrm{Fe}^{3+}$, so the content of $\mathrm{Cu}$ is obtained by $\mathrm{EDTA}$ titration.](1)</Checkpoint>

The amount of substance of $\mathrm{EDTA}$ consumed is $n_{\mathrm{EDTA}}=0.02035~\mathrm{mol/L}\times13.94~\mathrm{mL}=0.000283679~\mathrm{mol}$

$\mathrm{EDTA}$ and $\mathrm{Cu}^{2+}$ are complexed in a $1:1$ ratio, then in the titration solution $n_{\mathrm{Cu}}=n_{\mathrm{EDTA}}=0.000283679~\mathrm{mol}$

The amount of substance of copper in the alloy sample is $n_{\mathrm{Cu},\text{total}}=4n_{\mathrm{Cu}}=0.00113472~\mathrm{mol}$

The mass of copper in the alloy sample is $m_{\mathrm{Cu}} = 0.00113472 ~\mathrm{mol} \times 63.55 ~\mathrm{g/mol}= 0.0721112 ~\mathrm{g}$

The mass fraction of copper in the alloy sample is $\omega_{\mathrm{Cu}}=\dfrac{0.0721112 ~\mathrm{g}}{0.8527 \mathrm{g}}=8.45681~\%$

<Checkpoint>[$\omega_{\mathrm{Cu}}=8.457~\%$ ](0.5)</Checkpoint>

(c) $\mathrm{Fe}^{3+}$ is precipitated by ammonia and then dissolved in sulfuric acid, so the content of $\mathrm{Fe}$ is obtained by $\mathrm{EDTA}$ titration.

<Checkpoint>[(c) $\mathrm{Fe}^{3+}$ is precipitated by ammonia and then dissolved in sulfuric acid, so the content of $\mathrm{Fe}$ is obtained by $\mathrm{EDTA}$ titration.](1)</Checkpoint>

The amount of substance of $\mathrm{EDTA}$ consumed is $n_{\mathrm{EDTA}}=0.008140~\mathrm{mol/L}\times14.53~\mathrm{mL}=0.000118274~\mathrm{mol}$

$\mathrm{EDTA}$ and $\mathrm{Cu}^{2+}$ are complexed in a $1:1$ ratio, then in the titration solution $n_{\mathrm{Fe}}=n_{\mathrm{EDTA}}=0.000118274 ~\mathrm{mol}$

The amount of substance of iron in the alloy sample is $n_{\mathrm{Fe},\text{total}}=4\times10\times n_{\mathrm{Fe}}=0.00473097~\mathrm{mol}$

The mass of iron in the alloy sample is $m_{\mathrm{Fe}} = 0.00473097~\mathrm{mol} \times 55.85 ~\mathrm{g/mol}= 0.264225 ~\mathrm{g}$

The mass fraction of iron in the alloy sample is $\omega_{\mathrm{Fe}}=\dfrac{0.264225 ~\mathrm{g}}{0.8527 ~\mathrm{g}}=30.9868~\%$

<Checkpoint>[$\omega_{\mathrm{Fe}}=30.99~\%$ ](0.5)</Checkpoint>

$\dfrac{\omega_{\mathrm{Se}}}{\omega_{\mathrm{Fe}}}=\dfrac{55.2921~\%}{30.9868~\%}=1.78437$, so the mass fraction of $\mathrm{Se}$ is $1.784$ times that of $\mathrm{Fe}$, option K is incorrect.

<Checkpoint>[The mass fraction of $\mathrm{Se}$ is $1.784$ times that of $\mathrm{Fe}$, option K is incorrect](2)</Checkpoint>

Let the alloy be represented by $\mathrm{Fe}_x\mathrm{Cu}_y\mathrm{Se}$, then $x=\dfrac{30.9868~\%}{55.2921~\%}\times\dfrac{78.96~\mathrm{g/mol}}{55.85~\mathrm{g/mol}}=0.792315$

$y=\dfrac{8.45681~\%}{55.2921~\%}\times\dfrac{78.96~\mathrm{g/mol}}{63.55~\mathrm{g/mol}}=0.190036$

Thus $x+y=0.982351$, option I is incorrect.

<Checkpoint>[$x+y=0.982351$, option I is incorrect ](2)</Checkpoint>

Let the ratio of ferrous iron to ferric iron be $z:1$, then the valence balance is $0.792315\times\dfrac{2z+3}{z+1}+0.190036\times1+(-2)=0$, solving for $z=2.51618$
That is, the ratio of ferrous iron to ferric iron is approximately $2.516:1$, option J is correct.

<Checkpoint>[The ratio of ferrous iron to ferric iron is approximately $2.516:1$, option J is correct](2)</Checkpoint>

Therefore, choose J.

<Checkpoint>[Option J is correct](1)</Checkpoint>
抗坏血酸硫脲、氰离子、硫离子掩蔽 $\mathrm{Cu}$ 都会与 $\mathrm{Fe}^{3+}$ 反应，因此不能采用，故选项A, B, C均错误。

<Checkpoint>[因为抗坏血酸硫脲、氰离子、硫离子都与铁反应，所以选项A, B, C都错误](1)</Checkpoint>

过氧化氢在加热下分解、 $\mathrm{Fe}^{3+}$ 催化过氧化氢分解，消耗过氧化氢的量过多；且必须补充少量硝酸才能达到完全溶解样品的目的，因此不能替换，选项D错误。

<Checkpoint>[因为过氧化氢被铁催化、加热下分解，并且需要补加硝酸以溶解样品，选项D错误](1)</Checkpoint>

(a)实验中高温烘干沉淀将导致 $\mathrm{Se}$ 升华，获得的沉淀偏少，选项E错误

<Checkpoint>[高温烘干沉淀导致硒升华，损失沉淀，选项E错误](1)</Checkpoint>

(a)实验中如果没有保温一段时间，沉淀未充分陈化，获得的沉淀偏少，计算出的 $\mathrm{Se}$ 质量分数将变低，选项F错误。

<Checkpoint>[沉淀未完全陈化会损失沉淀，选项F错误](0.5)</Checkpoint>

<Checkpoint>[计算出的 $\mathrm{Se}$ 质量分数将变低，选项F错误](0.5)</Checkpoint>

(c)实验中加入氨水调整 $\mathrm{pH}$ 为微碱性，使 $\mathrm{Fe}$ 完全沉淀，微酸性将使获得的沉淀偏少，选项G错误。

<Checkpoint>[铁需要在微碱性环境下沉淀，选项G错误](0.5)</Checkpoint>

(c)实验中滴定温度为 $60-70$ 摄氏度，因为室温下 $\mathrm{EDTA}$和 $\mathrm{Fe}$ 的反应速度很慢，造成指示剂僵化，因此升温以提高 $\mathrm{EDTA}$和 $\mathrm{Fe}$ 的反应速度，避免指示剂僵化，选项H错误。

<Checkpoint>[室温下 $\mathrm{EDTA}$和 $\mathrm{Fe}$ 的反应速度很慢，造成指示剂僵化，选项H错误](1)</Checkpoint>

(a)中盐酸羟胺将 $\mathrm{Se}$ 沉淀，因此沉淀重量法测定得到 $\mathrm{Se}$ 的含量

<Checkpoint>[(a)中盐酸羟胺将 $\mathrm{Se}$ 沉淀](1)</Checkpoint>

合金样品中硒的重量 $m_{\mathrm{Se}}=0.2565~\mathrm{g}$ 

合金样品中硒的质量分数 $\omega_{\mathrm{Se}}=\dfrac{0.2565~\mathrm{g}}{0.4639~\mathrm{g}}=55.2921~\%$ 

<Checkpoint>[$\omega_{Se}=55.29~\%$ ](0.5)</Checkpoint>

(b)中氟化铵掩蔽了 $\mathrm{Fe}^{3+}$，因此 $\mathrm{EDTA}$ 滴定得到 $\mathrm{Cu}$ 的含量。

<Checkpoint>[(b)中氟化铵掩蔽了 $\mathrm{Fe}^{3+}$，因此 $\mathrm{EDTA}$ 滴定得到 $\mathrm{Cu}$ 的含量。](1)</Checkpoint>

消耗 $\mathrm{EDTA}$ 的物质的量 $n_{\mathrm{EDTA}}=0.02035~\mathrm{mol/L}\times13.94~\mathrm{mL}=0.000283679~\mathrm{mol}$ 

$\mathrm{EDTA}$ 与 $\mathrm{Cu}^{2+}$ 是 $1:1$ 络合，则滴定溶液中 $n_{\mathrm{Cu}}=n_{\mathrm{EDTA}}=0.000283679~\mathrm{mol}$ 

合金样品中铜的物质的量 $n_{\mathrm{Cu},\text{total}}=4n_{\mathrm{Cu}}=0.00113472~\mathrm{mol}$ 

合金样品中铜的质量 $m_{\mathrm{Cu}} = 0.00113472 ~\mathrm{mol} \times 63.55 ~\mathrm{g/mol}= 0.0721112 ~\mathrm{g}$ 

合金样品中铜的质量分数 $\omega_{\mathrm{Cu}}=\dfrac{0.0721112 ~\mathrm{g}}{0.8527 \mathrm{g}}=8.45681~\%$ 

<Checkpoint>[$\omega_{\mathrm{Cu}}=8.457~\%$ ](0.5)</Checkpoint>

(c)中 $\mathrm{Fe}^{3+}$ 经氨水沉淀再经硫酸溶解，因此 $\mathrm{EDTA}$ 滴定得到 $\mathrm{Fe}$ 的含量。

<Checkpoint>[(c)中 $\mathrm{Fe}^{3+}$ 经氨水沉淀再经硫酸溶解，因此 $\mathrm{EDTA}$ 滴定得到 $\mathrm{Fe}$ 的含量。](1)</Checkpoint>

消耗 $\mathrm{EDTA}$ 的物质的量 $n_{\mathrm{EDTA}}=0.008140~\mathrm{mol/L}\times14.53~\mathrm{mL}=0.000118274~\mathrm{mol}$ 

$\mathrm{EDTA}$ 与 $\mathrm{Cu}^{2+}$ 是 $1:1$ 络合，则滴定溶液中 $n_{\mathrm{Fe}}=n_{\mathrm{EDTA}}=0.000118274 ~\mathrm{mol}$ 

合金样品中铁的物质的量 $n_{\mathrm{Fe},\text{total}}=4\times10\times n_{\mathrm{Fe}}=0.00473097~\mathrm{mol}$ 

合金样品中铁的质量 $m_{\mathrm{Fe}} = 0.00473097~\mathrm{mol} \times 55.85 ~\mathrm{g/mol}= 0.264225 ~\mathrm{g}$ 

合金样品中铁的质量分数 $\omega_{\mathrm{Fe}}=\dfrac{0.264225 ~\mathrm{g}}{0.8527 ~\mathrm{g}}=30.9868~\%$ 

<Checkpoint>[$\omega_{\mathrm{Fe}}=30.99~\%$ ](0.5)</Checkpoint>

$\dfrac{\omega_{\mathrm{Se}}}{\omega_{\mathrm{Fe}}}=\dfrac{55.2921~\%}{30.9868~\%}=1.78437$，因此 $\mathrm{Se}$ 的质量分数是 $\mathrm{Fe}$ 的 $1.784$ 倍，选项K错误

<Checkpoint>[$\mathrm{Se}$ 的质量分数是 $\mathrm{Fe}$ 的 $1.784$ 倍，选项K错误](2)</Checkpoint>

以用 $\mathrm{Fe}_x\mathrm{Cu}_y\mathrm{Se}$ 表示合金，则 $x=\dfrac{30.9868~\%}{55.2921~\%}\times\dfrac{78.96~\mathrm{g/mol}}{55.85~\mathrm{g/mol}}=0.792315$

$y=\dfrac{8.45681~\%}{55.2921~\%}\times\dfrac{78.96~\mathrm{g/mol}}{63.55~\mathrm{g/mol}}=0.190036$ 

于是 $x+y=0.982351$ ，选项I错误

<Checkpoint>[$x+y=0.982351$，选项I错误 ](2)</Checkpoint>

设二价铁与三价铁的比例为 $z:1$，则价态平衡 $0.792315\times\dfrac{2z+3}{z+1}+0.190036\times1+(-2)=0$，解得 $z=2.51618$
也即二价铁与三价铁比例约 $2.516:1$，选项J正确。

<Checkpoint>[二价铁与三价铁比例约 $2.516:1$，选项J正确](2)</Checkpoint>

故选J。

<Checkpoint>[选项J正确](1)</Checkpoint>