Teacher Hui Hui obtained a white powder $\mathbf{A}$ in the laboratory one day. $\mathbf{A}$ is a binary compound containing the non-metallic element $\mathbf{X}$, which can quantitatively absorb carbon monoxide from the air and produce another gas present in the air along with the elemental form of $\mathbf{X}$. After dissolving $\mathbf{A}$ in cold water, an excess of sodium hydroxide solution was added to obtain a certain solution. When an excess of chlorine gas was passed through this solution, a colorless solution of $\mathbf{B}$ was obtained (Reaction 1), with excess sodium hypochlorite remaining in the solution. A pale pink hydrated crystal $\mathbf{C}$ (containing the metallic element $\mathbf{Y}$) was mixed with the aforementioned colorless solution. After thorough reaction, a bright red solution was obtained, and upon adding anhydrous ethanol, a hydrated crystal $\mathbf{D}$ precipitated. $\mathbf{D}$ contains no sulfur or chlorine elements.

To verify $\mathbf{D}$ (assuming it is a pure chemical composition), Teacher Hui conducted the following validation experiments:

1) A 0.5000 g sample of $\mathbf{D}$ was heated at 150°C until constant weight was achieved, leaving a residue of 0.3664 g.

2) Excess SO₂ gas was passed through a solution containing 0.5000 g of $\mathbf{D}$ (Reaction 2). After thorough reaction, the excess SO₂ was removed by heating, and the resulting solution was divided into two portions. To the first portion, a slight excess of silver nitrate solution was added, yielding 0.1452 g of a yellow precipitate $\mathbf{E}$. After centrifuging to remove $\mathbf{E}$, a slight excess of barium nitrate solution was added to the filtrate. The resulting precipitate was washed, dried, and weighed to yield 0.6252 g of a white precipitate $\mathbf{F}$. The second portion of the solution was thoroughly cooled, and a slight excess of $NH_4HS$ solution was added, producing a pink precipitate $\mathbf{G}$ containing the element $\mathbf{Y}$. $\mathbf{G}$ is a binary compound. After drying at $100\degree\mathrm{C}$ under a nitrogen atmosphere, the weight of $\mathbf{G}$ was measured as 0.0179 g.

Additional information:

1) The metallic element $\mathbf{Y}$ commonly occurs in nature as a binary compound $\mathbf{H}$, in which the mass fraction of $\mathbf{Y}$ is 63.2%.

2) The aqueous solution of $\mathbf{C}$ can react with barium chloride solution to also produce the white precipitate $\mathbf{F}$.

3) The mass fraction of element $\mathbf{X}$ in $\mathbf{B}$ is 43.18%, and the mass fraction of element $\mathbf{Y}$ in $\mathbf{C}$ is 19.84%.

Please deduce what $\mathbf{A, B, C, E, F, G, H, X, Y}$ are. Based on the answers, first calculate and deduce the chemical formula of $\mathbf{D}$, then deduce the chemical equations for Reactions 1 and 2 (not ionic equations). Select the appropriate options based on the results.
灰灰老师某天在实验室制得了一种白色粉末$\mathbf{A}$，$\mathbf{A}$是含有非金属元素$\mathbf{X}$的二元化合物，可以定量地吸收空气中的一氧化碳并得到空气中存在的另一种气体和$\mathbf{X}$的单质。将$\mathbf{A}$溶于冷水后得到某种溶液。向该溶液中加入过量的氢氧化钠溶液，并通入过量的氯气，得到$\mathbf{B}$的无色溶液（反应1），此时溶液中还存在过量的次氯酸钠。将某淡粉色结晶水合物$\mathbf{C}$（含有金属元素$\mathbf{Y}$）与上述无色溶液混合，充分作用后得到鲜红色溶液，加入无水乙醇，析出结晶水合物$\mathbf{D}$，$\mathbf{D}$中不含硫和氯元素。

为了验证$\mathbf{D}$（假设它是纯的化学组成），灰灰做了如下验证实验：

1）取0.5000g的样品$\mathbf{D}$，于150℃下加热至恒重，剩余固体的质量为0.3664g。

2）通过过量SO2气体于含有0.5000 g $\mathbf{D}$的溶液中（反应2），充分反应后加热除去过量的SO2，并将所得溶液分为两份。第1份中加入略过量的硝酸银溶液，得到0.1452 g的黄色沉淀$\mathbf{E}$，离心分离去掉沉淀$\mathbf{E}$后，向滤液中加入略过量的硝酸钡溶液，将所得沉淀洗涤并干燥后得到0.6252g白色沉淀$\mathbf{F}$。将第二份溶液充分冷却后加入稍过量的$NH_4HS$溶液，得到含有$\mathbf{Y}$元素的粉红色沉淀$\mathbf{G}$，$\mathbf{G}$是二元化合物。在$100\degree\mathrm{C}$的氮气氛围干燥后，称得$\mathbf{G}$的重量为0.0179g。

一些额外的信息：

1）金属元素$\mathbf{Y}$在自然界中在常见的存在形式是二元化合物$\mathbf{H}$，$\mathbf{H}$中$\mathbf{Y}$的质量分数为63.2%。

2）$\mathbf{C}$的水溶液可以与氯化钡溶液作用也产生白色沉淀$\mathbf{F}$。

3）$\mathbf{B}$中$\mathbf{X}$元素的质量分数为43.18%，$\mathbf{B}$不含结晶水。$\mathbf{C}$中$\mathbf{Y}$元素的质量分数为19.84%。

请推断$\mathbf{A, B, C, E, F, G, H, X, Y}$分别是什么，根据答案首先计算并推断$\mathbf{D}$的化学式，推断反应1、2的化学方程式（不是离子方程式）。根据结果选择最符合题意的唯一正确的选项。
The total number of atoms in the chemical formula of $\mathbf{A}$ is 3, the total number of atoms in the chemical formula of $\mathbf{B}$ is 10, and there are 5 mol of water of crystallization in 1 mol of $\mathbf{C}$.
The total number of atoms in chemical formula $\mathbf{A}$ is 4, the total number of atoms in chemical formula $\mathbf{B}$ is 11, and there are 6 mol of crystal water in 1 mol of $\mathbf{C}$.
The total number of atoms in chemical formula $\mathbf{A}$ is 5, the total number of atoms in chemical formula $\mathbf{B}$ is 12, and there are 7 mol of crystal water in 1 mol of $\mathbf{C}$.
The total number of atoms in chemical formula $\mathbf{A}$ is 6, the total number of atoms in chemical formula $\mathbf{B}$ is 13, and there are 8 mol of crystal water in 1 mol of $\mathbf{C}$.
$\mathbf{A}$ The total number of atoms in the chemical formula is 7.  

$\mathbf{B}$ The total number of atoms in the chemical formula is 14.  

1 mol $\mathbf{C}$ contains 9 mol of crystal water.
$\mathbf{D}$ is a relatively complex compound with a 1:1 molar ratio of $\mathbf{X}$ to $\mathbf{Y}$, containing fewer than 10 molecules of water of crystallization; the total number of metal element atoms in the chemical formula of $\mathbf{D}$ is 5.
$\mathbf{D}$ is a relatively complex compound with a molar ratio of $\mathbf{X}$ to $\mathbf{Y}$ of 2:1, containing fewer than 10 water molecules of crystallization; the total number of metal element atoms in the chemical formula of $\mathbf{D}$ is 6.
$\mathbf{D}$ is a relatively complex compound with a molar ratio of $\mathbf{X}$ to $\mathbf{Y}$ of 1:2, along with fewer than 10 water molecules of crystallization; the total number of metal element atoms in the chemical formula of $\mathbf{D}$ is 7.
$\mathbf{D}$ is a relatively complex compound, with a molar ratio of $\mathbf{X}$ to $\mathbf{Y}$ of 3:1, and containing fewer than 10 water molecules of crystallization; the total number of metal element atoms in the chemical formula of $\mathbf{D}$ is 9.
$\mathbf{D}$ is a relatively complex compound with a molar ratio of $\mathbf{X}$ to $\mathbf{Y}$ of 1:3, containing fewer than 10 water molecules of crystallization. The total number of metal element atoms in the chemical formula of $\mathbf{D}$ is 10.
$\mathbf{D}$ is a relatively complex compound in which the molar ratio of $\mathbf{X}$ to $\mathbf{Y}$ is 1:1, with more than 10 water molecules of crystallization; the total number of metal element atoms in the chemical formula of $\mathbf{D}$ is 5.
$\mathbf{D}$ is a relatively complex compound with a molar ratio of $\mathbf{X}$ to $\mathbf{Y}$ of 2:1, containing more than 10 water molecules of crystallization; the total number of metal element atoms in the chemical formula of $\mathbf{D}$ is 6.
$\mathbf{D}$ is a relatively complex compound with a molar ratio of $\mathbf{X}$ to $\mathbf{Y}$ of 1:2, containing more than 10 water molecules of crystallization; the total number of metal element atoms in the chemical formula of $\mathbf{D}$ is 7.
$\mathbf{D}$ is a relatively complex compound in which the molar ratio of $\mathbf{X}$ to $\mathbf{Y}$ is 2:2, with more than 10 water molecules of crystallization; the total number of metal element atoms in the chemical formula of $\mathbf{D}$ is 8.
$\mathbf{D}$ is a relatively complex compound, with a molar ratio of $\mathbf{X}$ to $\mathbf{Y}$ of 3:1, and more than 10 water molecules of crystallization; the total number of metal element atoms in the chemical formula of $\mathbf{D}$ is 9.
$\mathbf{D}$ is a relatively complex compound with a molar ratio of $\mathbf{X}$ to $\mathbf{Y}$ of 1:3, containing more than 10 molecules of water of crystallization; the total number of metal element atoms in the chemical formula of $\mathbf{D}$ is 10.
The product of the sum of coefficients on the left side and the sum of coefficients on the right side of Reaction 1 is 32, while the sum of the coefficients on the left side and the right side of Reaction 2 is 35.
The product of the sum of the coefficients on the left side and the sum of the coefficients on the right side of Reaction 1 is 35, and the sum of the coefficients on the left side and the right side of Reaction 2 is 32.
The product of the sum of coefficients on the left side and the sum of coefficients on the right side of reaction 1 is 35, while the sum of the coefficients on the left side and the right side of reaction 2 is 38.
The product of the sum of the coefficients on the left side and the sum of the coefficients on the right side of Reaction 1 is 38, and the sum of the coefficients on the left side and the right side of Reaction 2 is 35.
The product of the sum of the coefficients on the left side and the sum of the coefficients on the right side in Reaction 1 is 36. The sum of the coefficients on the left side and the right side in Reaction 2 is 40.
The product of the sum of coefficients on the left side and the sum of coefficients on the right side in Reaction 1 is 40, while the sum of the coefficients on the left side and the right side in Reaction 2 is 36.
Among the above options, the number of correct options is greater than 3.
Among the above options, the number of correct options is 2~3
None of the above options is correct
$\mathbf{A}$化学式中原子的总数为3，$\mathbf{B}$化学式中原子的总数为10，1 mol $\mathbf{C}$ 中有5 mol 结晶水。
$\mathbf{A}$化学式中原子的总数为4，$\mathbf{B}$化学式中原子的总数为11，1 mol $\mathbf{C}$ 中有6 mol 结晶水。
$\mathbf{A}$化学式中原子的总数为5，$\mathbf{B}$化学式中原子的总数为12，1 mol $\mathbf{C}$ 中有7 mol 结晶水。
$\mathbf{A}$化学式中原子的总数为6，$\mathbf{B}$化学式中原子的总数为13，1 mol $\mathbf{C}$ 中有8 mol 结晶水。
$\mathbf{A}$化学式中原子的总数为7，$\mathbf{B}$化学式中原子的总数为14，1 mol $\mathbf{C}$ 中有9 mol 结晶水。
$\mathbf{D}$ 是一个较为复杂的化合物，其中$\mathbf{X}$和$\mathbf{Y}$的摩尔比是1:1，还有少于10个结晶水；$\mathbf{D}$化学式中金属元素的原子总数为5。
$\mathbf{D}$ 是一个较为复杂的化合物，其中$\mathbf{X}$和$\mathbf{Y}$的摩尔比是2:1，还有少于10个结晶水；$\mathbf{D}$化学式中金属元素的原子总数为6。
$\mathbf{D}$ 是一个较为复杂的化合物，其中$\mathbf{X}$和$\mathbf{Y}$的摩尔比是1:2，还有少于10个结晶水；$\mathbf{D}$化学式中金属元素的原子总数为7。
$\mathbf{D}$ 是一个较为复杂的化合物，其中$\mathbf{X}$和$\mathbf{Y}$的摩尔比是3:1，还有少于10个结晶水；$\mathbf{D}$化学式中金属元素的原子总数为9。
$\mathbf{D}$ 是一个较为复杂的化合物，其中$\mathbf{X}$和$\mathbf{Y}$的摩尔比是1:3，还有少于10个结晶水；$\mathbf{D}$化学式中金属元素的原子总数为10。
$\mathbf{D}$ 是一个较为复杂的化合物，其中$\mathbf{X}$和$\mathbf{Y}$的摩尔比是1:1，还有多于10个结晶水；$\mathbf{D}$化学式中金属元素的原子总数为5。
$\mathbf{D}$ 是一个较为复杂的化合物，其中$\mathbf{X}$和$\mathbf{Y}$的摩尔比是2:1，还有多于10个结晶水；$\mathbf{D}$化学式中金属元素的原子总数为6。
$\mathbf{D}$ 是一个较为复杂的化合物，其中$\mathbf{X}$和$\mathbf{Y}$的摩尔比是1:2，还有多于10个结晶水；$\mathbf{D}$化学式中金属元素的原子总数为7。
$\mathbf{D}$ 是一个较为复杂的化合物，其中$\mathbf{X}$和$\mathbf{Y}$的摩尔比是2:2，还有多于10个结晶水；$\mathbf{D}$化学式中金属元素的原子总数为8。
$\mathbf{D}$ 是一个较为复杂的化合物，其中$\mathbf{X}$和$\mathbf{Y}$的摩尔比是3:1，还有多于10个结晶水；$\mathbf{D}$化学式中金属元素的原子总数为9。
$\mathbf{D}$ 是一个较为复杂的化合物，其中$\mathbf{X}$和$\mathbf{Y}$的摩尔比是1:3，还有多于10个结晶水；$\mathbf{D}$化学式中金属元素的原子总数为10。
反应1左侧系数之和和右侧系数之和的乘积是32，反应2左侧和右侧系数之和的加和是35。
反应1左侧系数之和和右侧系数之和的乘积是35，反应2左侧和右侧系数之和的加和是32。
反应1左侧系数之和和右侧系数之和的乘积是35，反应2左侧和右侧系数之和的加和是38。
反应1左侧系数之和和右侧系数之和的乘积是38，反应2左侧和右侧系数之和的加和是35。
反应1左侧系数之和和右侧系数之和的乘积是36，反应2左侧和右侧系数之和的加和是40。
反应1左侧系数之和和右侧系数之和的乘积是40，反应2左侧和右侧系数之和的加和是36。
以上选项中，正确的选项数量大于3个
以上选项中，正确的选项数量为2~3
以上选项中不存在正确选项
First, we must deduce both the elements and compounds:

Based on the description, A is a white powder of a binary compound containing nonmetal element X, which can quantitatively absorb carbon monoxide and produce another gas in the air along with the elemental form of X. We know that the classic white reagent used to quantitatively absorb CO is iodine pentoxide $I_2O_5$, meaning $\mathbf{A}$ is $I_2O_5$.

<Checkpoint>[$\mathbf{A}$ is $I_2O_5$](0.5)</Checkpoint>

Dissolving $\mathbf{A}$ in cold water produces $HIO_3$, which is neutralized by excess $NaOH$ and oxidized under alkaline conditions to form periodate. The relative atomic mass of iodine is 126.90, and based on the mass fraction of 43.18%, the chemical formula of $\mathbf{B}$ should be $Na_3H_2IO_6$.

<Checkpoint>[The chemical formula of $\mathbf{B}$ is $Na_3H_2IO_6$](1)</Checkpoint>

We know that $\mathbf{C}$ is a pale pink crystalline hydrate containing metal element $\mathbf{Y}$. From the pale pink color, we can infer that it is likely a hydrate containing divalent manganese ions $Mn^{2+}$. The solution of $\mathbf{C}$ reacts with barium chloride to form a white precipitate, indicating the presence of sulfate ions. Thus, the white precipitate $\mathbf{F}$ is $BaSO_4$, and $\mathbf{C}$ should be $MnSO_4\cdot nH_2O$. Based on the mass fraction, $n\approx 7$, so we can deduce that $\mathbf{Y}$ is $Mn$.

<Checkpoint>[$\mathbf{C}$ is $MnSO_4\cdot 7H_2O$](1)</Checkpoint>

$SO_2$ is a strongly reducing gas. Passing it through a solution containing D should reduce the high-valent manganese ions and periodate ions. Adding silver nitrate solution to the reduced solution yields a yellow precipitate $\mathbf{E}$. Given the properties of iodide and silver ions, $\mathbf{E}$ is confirmed to be $AgI$.

<Checkpoint>[$\mathbf{E}$ is $AgI$](0.5)</Checkpoint>

In another portion of the solution, adding NH₄HS produces a pink precipitate $\mathbf{G}$ containing $Mn$. Manganese ions react with sulfide or hydrosulfide ions to form pink manganese sulfide, meaning $\mathbf{G}$ is $MnS$.

<Checkpoint>[$\mathbf{G}$ is $MnS$](0.5)</Checkpoint>

Next, we should analyze how to determine the chemical formula of $\mathbf{D}$:

Based on the reactants that produce $\mathbf{D}$, $\mathbf{D}$ should contain high-valent manganese, periodate ions, and water of crystallization. Under the influence of excess $SO_2$, the high-valent manganese is reduced to $Mn^{2+}$, and the periodate ions are reduced to $I^-$.

In 0.2500 g of $\mathbf{D}$:

n($I$) = n($AgI$) = 0.1452/234.8 = 6.184×10⁻⁴(mol)

<Checkpoint>[Calculated n($I$) = 6.184×10⁻⁴(mol)](0.5)</Checkpoint>

n($Mn$) = n($MnS$) = 0.0179/87.00 = 2.06×10⁻⁴(mol)

<Checkpoint>[Calculated n($Mn$) = 2.06×10⁻⁴(mol)](0.5)</Checkpoint>

Total sulfate ions produced by the oxidation of $SO_2$:

n($SO_4^{2-}$) = n($BaSO_4$) = 0.6252/233.4 = 2.679×10⁻³(mol)

<Checkpoint>[Calculated n($SO_4^{2-}$) = 2.679×10⁻³(mol)](0.5)</Checkpoint>

It follows that n($Mn$) : n($I$) : n($SO_4^{2-}$) = 1:3:13

<Checkpoint>[Deduced n($Mn$) : n($I$) : n($SO_4^{2-}$) = 1:3:13](1)</Checkpoint>

From the conservation of electrons in the redox reaction:

13$SO_2$ → 13$SO_4^{2-}$ ~ loses 26$e^-$

3$I$(VII) → 3$I^-$ gains 24$e^-$

Thus, $Mn$(X) → $Mn$(II) ~ should gain 2$e^-$

Therefore, the oxidation state of $Mn$ in $\mathbf{D}$ is +4, i.e., X=+4

<Checkpoint>[Deduced the oxidation state of $Mn$ in $\mathbf{D}$ is +4](2)</Checkpoint>

At this point, it is clear that $\mathbf{D}$ contains 1 $Mn$, 3 $IO_6^{5-}$, and several water molecules. Due to the significant number of anions, $\mathbf{D}$ should also contain several $Na^+$ or $H^+$ to balance the charge. (These $H^+$ may be bound to the periodate ions.)

Assume the chemical formula of $\mathbf{D}$ is $Na_xH_yMn(IO_6)_3 \cdot z H_2O$.

Then M = 0.2500/(2.06×10⁻⁴) = 1213(g/mol)

<Checkpoint>[Calculated the relative atomic mass of $\mathbf{D}$ as 1213(g/mol)](1)</Checkpoint>

M($H_2O$) = 1213×(0.5000 - 0.3664)/0.5000 = 324.3(g/mol), so z = 18

<Checkpoint>[Calculated there should be 18 water molecules of crystallization](1)</Checkpoint>

From charge conservation, $x + y + 4 = 3\times5$ gives y = 11 - x.

Also, $22.99x + 54.94 + 3\times222.7 + 1.008(11-x) + 324.3 = 1213$.

Solving gives x = 7, so y = 4.

<Checkpoint>[Calculated $\mathbf{D}$ contains 7 $Na^+$](1)</Checkpoint>

<Checkpoint>[Calculated $\mathbf{D}$ contains 4 $H^+$](1)</Checkpoint>

Thus, the chemical formula of $\mathbf{D}$ is $Na_7H_4Mn(IO_6)_3\cdot 18H₂O$.

<Checkpoint>[Deduced the chemical formula of $\mathbf{D}$ is $Na_7H_4Mn(IO_6)_3\cdot 18H₂O$](0.5)</Checkpoint>

Finally, we can write reactions 1 and 2.

Reaction 1: $HIO_3 + 5NaOH + Cl_2 \rightarrow Na_3H_2IO_6 + 2NaCl + 2H₂O$

<Checkpoint>[Wrote reaction 1: $HIO_3 + 5NaOH + Cl_2 \rightarrow Na_3H_2IO_6 + 2NaCl + 2H₂O$](1)</Checkpoint>

Reaction 2: $Na_7H_4Mn(IO_6)_3 + 13SO_2 + 8H_2O → 3NaI + MnSO_4 + 2Na_2SO_4 + 10 H_2SO_4$

<Checkpoint>[Wrote reaction 2: $Na_7H_4Mn(IO_6)_3 + 13SO_2 + 8H_2O → 3NaI + MnSO_4 + 2Na_2SO_4 + 10 H_2SO_4$](1)</Checkpoint>

We know that the sum of coefficients on the left side of reaction 1 is 7, and on the right side is 5. For reaction 2, the left side sums to 22, and the right side sums to 16. The product of coefficients for reaction 1 is 35, and for reaction 2, it is 38.
首先，我们必须把元素和化合物都推断出来：

根据描述，A是一种含有非金属元素X的二元化合物白色粉末，可以定量吸收一氧化碳，并生成空气中的另一种气体和X的单质。我们知道，用来定量吸收$CO$的经典的白色试剂是五氧化二碘$I_2O_5$，也就是说$\mathbf {A}$是$I_2O_5$，X为$I$

<Checkpoint>[$\mathbf {A}$是$I_2O_5$](0.5)</Checkpoint>

将$\mathbf{A}$溶于冷水生成$HIO_3$，加入过量的$NaOH$后被中和，并在碱性条件下被氧化，得到高碘酸盐。我们知道碘的相对原子质量是 126.90, 根据质量分数为43.18 %，可以计算出$\mathbf{B}$的化学式应当为$Na_3H_2IO_6$。

<Checkpoint>[$\mathbf{B}$的化学式是$Na_3H_2IO_6$](1)</Checkpoint>

我们知道，$\mathbf{C}$是某种淡粉色的结晶水合物，其中含有金属元素$\mathbf{Y}$。从淡粉色我们可以推测出应当是含有二价锰离子$Mn^{2+}$的某种水合物。利用提示金属元素$\mathbf{Y}$在自然界中在常见的存在形式是二元化合物$\mathbf{H}$，$\mathbf{H}$中$\mathbf{Y}$的质量分数为63.2%。可以验证$\mathbf{H}$为$MnO_2$。$\mathbf{C}$的溶液能够和氯化钡生成白色沉淀，可以推断出应当含有硫酸根离子，那么白色沉淀$\mathbf{F}$是$BaSO_4$，并且$\mathbf{C}$应当是$MnSO_4\cdot nH_2O$。根据质量分数可以计算出$n\approx 7$，那么可以推断出，$\mathbf{Y}$是$Mn$。

<Checkpoint>[$\mathbf{C}$是$MnSO_4\cdot 7H_2O$](1)</Checkpoint>


$SO_2$是有强还原性的气体，将气体通入含有D的溶液中应当会还原其中的高价锰离子和高碘酸根。还原后的溶液中加入硝酸银溶液，得到黄色沉淀$\mathbf{E}$，由碘离子和银离子的性质可以推出其正是$AgI$。

<Checkpoint>[$\mathbf{E}$是$AgI$](0.5)</Checkpoint>

另一份溶液中加入NH₄HS，得到含有$Mn$的粉红色沉淀$\mathbf{G}$。锰离子与硫离子或硫氢根离子反应生成粉红色的硫化锰，也就是$\mathbf{G}$即为 $MnS$。

<Checkpoint>[$\mathbf{G}$是 $MnS$](0.5)</Checkpoint>

然后，我们应当开始分析如何得到$\mathbf{D}$的化学式：

根据产生$\mathbf{D}$的反应物，$\mathbf{D}$中应含有高价锰，高碘酸根和结晶水。故在过量$SO_2$的作用下，高价锰被还原成$Mn^{2+}$，高碘酸根被还原成$I^-$。

在0.2500 g $\mathbf{D}$中：

n($I$) = n($AgI$) = 0.1452/234.8 = 6.184×10⁻⁴(mol) 

<Checkpoint>[计算出n($I$) = 6.184×10⁻⁴(mol) ](0.5)</Checkpoint>

n($Mn$) = n($MnS$) = 0.0179/87.00 = 2.06×10⁻⁴mol) 

<Checkpoint>[计算出 n($Mn$) = 2.06×10⁻⁴mol ](0.5)</Checkpoint>

$SO_2$被氧化产生的硫酸根总量：

n($SO_4^{2-}$) = n($BaSO_4$) = 0.6252/233.4 = 2.679×10⁻³(mol) 

<Checkpoint>[计算出 n($SO_4^{2-}$) =  2.679×10⁻³ (mol) ](0.5)</Checkpoint>

易知n($Mn$) : n($I$) : n($SO_4^{2-}$) = 1:3:13

<Checkpoint>[推出 n($Mn$) : n($I$) : n($SO_4^{2-}$) = 1:3:13](1)</Checkpoint>

由氧化还原反应的得失电子守恒，

13$SO_2$ → 13$SO_4^{2-}$ ~ 失去 26$e^-$

3$I$(VII) → 3$I^-$ 得到 24$e^-$

故 $Mn$(X) → $Mn$(II) ~ 应得到 2$e^-$

故可知 $\mathbf{D}$ 中的 $Mn$ 氧化态为 +4 即 X=+4

<Checkpoint>[推出$\mathbf{D}$ 中的 $Mn$ 氧化态为 +4 ](2)</Checkpoint>

至此，明确已知 $\mathbf{D} $中含有 1 个 $Mn$，3 个 $IO_6^{5-}$，若干个结晶水。由于负离子显著较多，所以 D 中应该还含有若干个 $Na^+$或者 $H^+$来平衡电荷。（这些 $H^+$可以结合在高碘酸根上）

可设 D 的化学式为 $Na_xH_yMn(IO_6)_3 \cdot z H_2O$ 

则 M = 0.2500/(2.06×10⁻⁴) = 1213(g/mol) 

<Checkpoint>[计算出$\mathbf{D}$的相对原子质量为1213(g/mol) ](1)</Checkpoint>

M($H_2O$) = 1213×(0.5000 - 0.3664)/0.5000 = 324.3(g/mol)，则 z = 18 

<Checkpoint>[计算出应当有18个结晶水](1)</Checkpoint>

根据电荷守恒，$x + y + 4 = 3\times5$ 可得 y = 11 - x 

又 $22.99x + 54.94 + 3\times222.7 + 1.008(11-x) + 324.3 = 1213$

解得 x = 7，则 y = 4

<Checkpoint>[计算出$\mathbf{D}$中$Na^+$有7个](1)</Checkpoint>

<Checkpoint>[计算出$\mathbf{D}$中$H^+$有4个](1)</Checkpoint>

故 $\mathbf{D}$ 的化学式为 $Na_7H_4Mn(IO_6)_3\cdot 18H₂O$ 

<Checkpoint>[推出$\mathbf{D}$ 的化学式为 $Na_7H_4Mn(IO_6)_3\cdot 18H₂O$ ](0.5)</Checkpoint>

最后，我们可以写出反应1，2分别是什么。

反应 1：$HIO_3 + 5NaOH + Cl_2 \rightarrow Na_3H_2IO_6 + 2NaCl + 2H₂O$

<Checkpoint>[写出反应1：$HIO_3 + 5NaOH + Cl_2 \rightarrow Na_3H_2IO_6 + 2NaCl + 2H₂O$](1)</Checkpoint>

反应 2： $Na_7H_4Mn(IO_6)_3 + 13SO_2 + 8H_2O → 3NaI + MnSO_4 + 2Na_2SO_4 + 10 H_2SO_4$

<Checkpoint>[写出反应 2： $Na_7H_4Mn(IO_6)_3 + 13SO_2 + 8H_2O → 3NaI + MnSO_4 + 2Na_2SO_4 + 10 H_2SO_4$](1)</Checkpoint>

我们知道，反应1左侧的系数之和是7，右侧系数之和是5；反应2左侧的系数之和是22，右侧的系数之和是16。可以算出反应1两侧系数之积是35，反应2两侧系数之和是38.