The basic process of crude salt purification is as follows:

Dissolve $10~g$ of crude salt in $40~mL$ of deionized water with heating, then add $\mathrm{25\%~BaCl_2}$ solution to it until no more precipitate is produced, keep warm for $10~min$, and perform the first normal pressure filtration. Then add saturated $\mathrm{Na_2CO_3}$ solution until no more precipitate is produced, and perform the second normal pressure filtration. Add 6 M hydrochloric acid, then evaporate by heating in a water bath, filter under suction to remove the mother liquor, and finally dry by stir-frying to obtain refined salt.

Assume that $10~g$ of crude salt is composed of $9~g$ $\mathrm{NaCl}$, $0.3~g$ $\mathrm{MgCl_2}$, $0.3~g$ $\mathrm{CaCl_2}$, $0.2~g$ $\mathrm{FeCl_3}$ and $0.2~g$ $\mathrm{Na_2SO_4}$. It is known that $\mathrm{BaCl_2}$ and $\mathrm{Na_2CO_3}$ are added with a 50% excess, $1~mL$ of solution is lost in each normal pressure filtration, and $5~mL$ of mother liquor is lost in vacuum filtration. Estimate the theoretical yield and theoretical yield rate based on the following solubility data.

Hint: It is considered that the volume of the solution remains unchanged and remains at $40~mL$ before evaporation and concentration, that is, deionized water will be added after the volume decreases (such as after normal pressure filtration), and the excess water will evaporate after the volume increases; vacuum filtration is performed immediately after evaporation and concentration, and it can be considered that the temperature at this time is about the boiling point of water; it is considered that $1~mL$ of aqueous solution contains $1~mL$ of water.

The data in the table below is the mass fraction of solute in saturated $\mathrm{NaCl}$ solution.

|Temperature|Mass Fraction|
|---|---|
|$25\degree C$|26.45%|
|$100\degree C$|28.05%|
粗盐提纯的基本流程如下：

用$40~mL$去离子水加热溶解$10~g$粗盐，然后向其中加入$\mathrm{25\%~BaCl_2}$溶液至不再产生沉淀，保温$10~min$，第一次常压过滤。然后加入饱和 $\mathrm{Na_2CO_3}$ 溶液至不再产生沉淀，第二次常压过滤。加入6 M盐酸，然后水浴加热蒸发，抽滤去除母液，最后炒干，得到精盐。

假定$10~g$粗盐是由$9~g$ $\mathrm{NaCl}$，$0.3~g$ $\mathrm{MgCl_2}$，$0.3~g$ $\mathrm{CaCl_2}$，$0.2~g$ $\mathrm{FeCl_3}$与$0.2~g$ $\mathrm{Na_2SO_4}$混合而成。已知加入 $\mathrm{BaCl_2}$和$\mathrm{Na_2CO_3}$时均会过量50% ，一次常压过滤损失$1~mL$溶液，减压过滤损失$5~mL$母液，根据下列溶解度数据，估算理论产量和理论产率。

提示：认为在蒸发浓缩之前，溶液的体积不变并保持在$40~mL$，即体积减少（如在常压过滤完成后）后会补充去离子水，体积增加后多余的水会挥发；减压过滤是在蒸发浓缩后立即进行的，可以认为此时温度约为水的沸点；认为$1~mL$水溶液中含有$1~mL$水。

下表中数据为$\mathrm{NaCl}$饱和溶液中溶质的质量分数。

|温度|质量分数|
|---|---|
|$25\degree C$|26.45%|
|$100\degree C$|28.05%|
All other options are incorrect
$5.69~g$, 56.9%
$6.24~g$, 62.4%
$7.70~g$, 77.0%
$8.16~g$, 81.6%
$8.54~g$, 85.4%
$9.11~g$, 91.1%
$5.94~g$, 66.0%
$6.73~g$, 74.8%
$7.32~g$, 81.3%
$7.92~g$, 88.0%
$8.34~g$, 92.7%
$8.70~g$, 96.7%
$9.33~g$, 103.7%
其他选项均不正确
$5.69~g$，56.9%
$6.24~g$，62.4%
$7.70~g$，77.0%
$8.16~g$，81.6%
$8.54~g$，85.4%
$9.11~g$，91.1%
$5.94~g$，66.0%
$6.73~g$，74.8%
$7.32~g$，81.3%
$7.92~g$，88.0%
$8.34~g$，92.7%
$8.70~g$，96.7%
$9.33~g$，103.7%
The composition of the initial solution is shown in the following table:

||$\mathrm{Na^+}$|$\mathrm{Cl^-}$|$\mathrm{Mg^{2+}}$|$\mathrm{Ca^{2+}}$|$\mathrm{Fe^{3+}}$|$\mathrm{SO_4^{2-}}$|$\mathrm{Ba^{2+}}$|
|---|---|---|---|---|---|---|---|
|n/mmol|156.8|169.4|3.151|2.703|1.233|1.408|0|

After adding an excess of 50% $\mathrm{BaCl_2}$, $\mathrm{SO_4^{2-}}$ is converted to $\mathrm{BaSO_4}$.

||$\mathrm{Na^+}$|$\mathrm{Cl^-}$|$\mathrm{Mg^{2+}}$|$\mathrm{Ca^{2+}}$|$\mathrm{Fe^{3+}}$|$\mathrm{SO_4^{2-}}$|$\mathrm{Ba^{2+}}$|
|---|---|---|---|---|---|---|---|
|n/mmol|156.8|173.6|3.151|2.703|1.233|0|0.704|

<Checkpoint>[The amounts of substance of $\mathrm{Cl^-}$ and $\mathrm{Ba^{2+}}$ after adding an excess of 50% $\mathrm{BaCl_2}$ are $173.6~mmol$ and $0.704~mmol$, respectively.](1)</Checkpoint>

After filtration with a loss of 1/40, the composition becomes:

||$\mathrm{Na^+}$|$\mathrm{Cl^-}$|$\mathrm{Mg^{2+}}$|$\mathrm{Ca^{2+}}$|$\mathrm{Fe^{3+}}$|$\mathrm{SO_4^{2-}}$|$\mathrm{Ba^{2+}}$|
|---|---|---|---|---|---|---|---|
|n/mmol|152.9|169.3|3.072|2.635|1.202|0|0.686|

Then, an excess of 50% $\mathrm{Na_2CO_3}$ was added. First, write out the equations for the reactions of each ion:

$\mathrm{Fe^{3+}+3CO_3^{2-}+3H_2O\to Fe(OH)_3+3HCO_3^-}$

<Checkpoint>[$\mathrm{Fe^{3+}+3CO_3^{2-}+3H_2O\to Fe(OH)_3+3HCO_3^-}$](0.5)</Checkpoint>

$\mathrm{Ca^{2+}+CO_3^{2-}\to CaCO_3}$

$\mathrm{2Mg^{2+}+3CO_3^{2-}+2H_2O\to Mg_2(OH)_2CO_3+2HCO_3^-}$

<Checkpoint>[$\mathrm{2Mg^{2+}+3CO_3^{2-}+2H_2O\to Mg_2(OH)_2CO_3+2HCO_3^-}$](0.5)</Checkpoint>

$\mathrm{Ba^{2+}+CO_3^{2-}\to BaCO_3}$

Therefore, the stoichiometric ratio is $\mathrm{1Fe^{3+}-\frac{3}{2}CO_3^{2-},1Ca^{2+}-1CO_3^{2-},1Mg^{2+}-1CO_3^{2-},1Ba^{2+}-1CO_3^{2-}}$

At this point, only focus on $\mathrm{Na^+}$ in the reaction. After adding $\mathrm{Na_2CO_3}$, all metal cations in the solution except $\mathrm{Na^+}$ are converted to the corresponding precipitates, but the total amount of charge carried by their anions remains unchanged. Therefore, the amount of substance of cations can also be determined by charge conservation. Then, according to the stoichiometric ratio, $\mathrm{Na^+}$ in the solution is: $152.9+(2\times3.072+2\times2.635+3\times1.202+2\times0.686)\times1.5=177.5~mmol$

<Checkpoint>[The amount of $\mathrm{Na^+}$ in the solution after adding $\mathrm{Na_2CO_3}$ is $177.5~mmol$](1)</Checkpoint>

Then, after filtration with a loss of 1/40 and addition of concentrated hydrochloric acid, $173.1~mmol$ $\mathrm{NaCl}$ is obtained, which is $10.11~g$ $\mathrm{NaCl}$.

<Checkpoint>[There is $10.11~g $ $\mathrm{NaCl}$ in the solution after adding concentrated hydrochloric acid](1)</Checkpoint>

Then, vacuum filtration loses $5~mL$ of saturated $\mathrm{NaCl}$ solution, and the mass of $\mathrm{NaCl}$ in it is $5~g×\frac{28.05\%}{1-28.05\%}\approx1.95~g$.

<Checkpoint>[$1.95~g$ $\mathrm{NaCl}$ is lost by vacuum filtration](1)</Checkpoint>

Therefore, the final theoretical yield is $8.16~g$, and the theoretical yield is 81.6%.
最初溶液的组成如下表：

||$\mathrm{Na^+}$|$\mathrm{Cl^-}$|$\mathrm{Mg^{2+}}$|$\mathrm{Ca^{2+}}$|$\mathrm{Fe^{3+}}$|$\mathrm{SO_4^{2-}}$|$\mathrm{Ba^{2+}}$|
|---|---|---|---|---|---|---|---|
|n/mmol|156.8|169.4|3.151|2.703|1.233|1.408|0|

加入过量50%的$\mathrm{BaCl_2}$后$\mathrm{SO_4^{2-}}$转化为$\mathrm{BaSO_4}$。

||$\mathrm{Na^+}$|$\mathrm{Cl^-}$|$\mathrm{Mg^{2+}}$|$\mathrm{Ca^{2+}}$|$\mathrm{Fe^{3+}}$|$\mathrm{SO_4^{2-}}$|$\mathrm{Ba^{2+}}$|
|---|---|---|---|---|---|---|---|
|n/mmol|156.8|173.6|3.151|2.703|1.233|0|0.704|

<Checkpoint>[加入过量50%的$\mathrm{BaCl_2}$后$\mathrm{Cl^-}$和$\mathrm{Ba^{2+}}$的物质量分别为$173.6~mmol$和$0.704~mmol$](1)</Checkpoint>

之后过滤损失1/40，组成变为：

||$\mathrm{Na^+}$|$\mathrm{Cl^-}$|$\mathrm{Mg^{2+}}$|$\mathrm{Ca^{2+}}$|$\mathrm{Fe^{3+}}$|$\mathrm{SO_4^{2-}}$|$\mathrm{Ba^{2+}}$|
|---|---|---|---|---|---|---|---|
|n/mmol|152.9|169.3|3.072|2.635|1.202|0|0.686|

然后加入了过量50%的$\mathrm{Na_2CO_3}$，先写出各个离子反应的方程式：

$\mathrm{Fe^{3+}+3CO_3^{2-}+3H_2O\to Fe(OH)_3+3HCO_3^-}$

<Checkpoint>[$\mathrm{Fe^{3+}+3CO_3^{2-}+3H_2O\to Fe(OH)_3+3HCO_3^-}$](0.5)</Checkpoint>

$\mathrm{Ca^{2+}+CO_3^{2-}\to CaCO_3}$

$\mathrm{2Mg^{2+}+3CO_3^{2-}+2H_2O\to Mg_2(OH)_2CO_3+2HCO_3^-}$

<Checkpoint>[$\mathrm{2Mg^{2+}+3CO_3^{2-}+2H_2O\to Mg_2(OH)_2CO_3+2HCO_3^-}$](0.5)</Checkpoint>

$\mathrm{Ba^{2+}+CO_3^{2-}\to BaCO_3}$

因此化学计量比为$\mathrm{1Fe^{3+}-\frac{3}{2}CO_3^{2-},1Ca^{2+}-1CO_3^{2-},1Mg^{2+}-1CO_3^{2-},1Ba^{2+}-1CO_3^{2-}}$

此时只需要关注反应中的$\mathrm{Na^+}$即可，加入$\mathrm{Na_2CO_3}$后溶液中除$\mathrm{Na^+}$之外的金属阳离子全部转化为相应的沉淀，但其阴离子所带的总电荷量是不变的，因此阳离子的物质的量也可以有电荷守恒来确定，则根据计量比，溶液中$\mathrm{Na^+}$为：$152.9+(2\times3.072+2\times2.635+3\times1.202+2\times0.686)\times1.5=177.5~mmol$

<Checkpoint>[加入$\mathrm{Na_2CO_3}$后溶液中$\mathrm{Na^+}$为$177.5~mmol$](1)</Checkpoint>

然后过滤损失1/40并加入浓盐酸，得到$173.1~mmol$ $\mathrm{NaCl}$，即为$10.11~g$ $\mathrm{NaCl}$。

<Checkpoint>[加入浓盐酸后溶液中有$10.11~g $ $\mathrm{NaCl}$](1)</Checkpoint>

然后抽滤损失$5~mL$饱和$\mathrm{NaCl}$溶液，其中的$\mathrm{NaCl}$质量为$5~g×\frac{28.05\%}{1-28.05\%}\approx1.95~g$。

<Checkpoint>[抽滤损失$1.95~g$ $\mathrm{NaCl}$](1)</Checkpoint>

因此最终理论产量为$8.16~g$，理论产率为81.6%。