$\rm CuSO_4$, $\rm H_5IO_6$, $\rm K_2S_2O_8$, and $\rm KOH$ were mixed in a certain proportion, $\rm 200~mL$ of distilled water was added, heated to boiling for $\rm 30~min$, cooled, and filtered to obtain a blue-green solution (Reaction 1); an appropriate amount of $\rm 10\%~NaOH$ solution was added to it, allowed to stand for $\rm 48~h$, and blue-green crystals precipitated. Partial analysis results of the blue-green crystals are as follows:

(1) X-ray diffraction analysis showed that the crystal is a complex salt with crystal water, the cations are $\rm Na^+$ and $\rm K^+$, and the anion is a mononuclear coordination ion centered on copper, with $\rm HIO_6^{4-}$ as the ligand.

(2) $\rm 0.2001~g$ of the crystal was weighed, $\rm 10~mL$ of hydrochloric acid was added, and heated to dissolve the sample, then diluted with $\rm 50~mL$ of distilled water, buffer solution and $\rm PAN$ indicator were added, and titrated with $\rm 0.01973~mol\cdot L^{-1}~EDTA$ solution, consuming $\rm 11.81~mL$.

(3) $\rm 0.2059~g$ of the crystal was weighed, $\rm 10~mL~25\%~KI$ solution and $\rm 10~mL~25\%$ sulfuric acid solution were added (Reaction 2), after full reaction, with starch as an indicator, titrated with $\rm 0.1024~mol\cdot L^{-1}$ sodium thiosulfate solution, consuming $\rm 42.16~mL$.

(4) Thermogravimetric analysis of the crystal showed a weight loss of $\rm 25.18\%$, corresponding to the loss of all crystal water.

Which of the following statements are correct:

1.  $\rm Cu$ in the crystal is +5 valence
2.  The chemical formula of the crystal contains 12 crystal water molecules
3.  Without considering crystal water, the sum of the left and right coefficients of the chemical equation of Reaction 1 is 45
4.  Without considering crystal water, the difference between the left and right coefficients of the chemical equation of Reaction 2 is 6 (right minus left)
将$\rm CuSO_4$、$\rm H_5IO_6$、$\rm K_2S_2O_8$和$\rm KOH$按一定比例混合，加入蒸馏水$\rm 200~mL$，加热煮沸$\rm 30~min$，冷却，过滤，得蓝绿色溶液（反应1）；向其中加入适量$\rm 10\%~NaOH$ 溶液，静置$\rm 48~h$，析出蓝绿色晶体，对该蓝绿色晶体的部分分析结果如下：

（1）X-射线衍射分析结果表明，该晶体为一种带结晶水的复盐，阳离子为$\rm Na^+$和$\rm K^+$，阴离子为以铜为中心的单核配离子，配体为$\rm HIO_6^{4-}$。

（2）称取该晶体$\rm 0.2001~g$，加入$\rm 10~mL$盐酸，加热以溶解样品，随后加入$\rm 50~mL$蒸馏水稀释，加入缓冲液和$\rm PAN$指示剂，用$\rm 0.01973~mol\cdot L^{-1}~EDTA$溶液滴定，消耗$\rm 11.81~mL$。

（3）称取该晶体$\rm 0.2059~g$，加入$\rm 10~mL~25\%~KI$溶液和$\rm 10~mL~25\%$硫酸溶液（反应2），充分反应后，以淀粉为指示剂，用$\rm 0.1024~mol\cdot L^{-1}$硫代硫酸钠溶液滴定，消耗$\rm 42.16~mL$。

（4）取该晶体进行热重分析，失重$\rm 25.18\%$，对应于失去所有结晶水。

下列说法正确的有：

1. 该晶体中$\rm Cu$为+5价
2. 该晶体的化学式中含12个结晶水
3. 不考虑结晶水时，反应1的化学方程式左右系数之和为45
4. 不考虑结晶水时，反应2的化学方程式左右系数之差为6(右减左)
All other options are incorrect
1
2
3
4
1,2
1,3
1,4
2,3
2,4
3,4
1,2,3
1,2,4
1,3,4
2,3,4
1,2,3,4
其他选项均不正确
1
2
3
4
1,2
1,3
1,4
2,3
2,4
3,4
1,2,3
1,2,4
1,3,4
2,3,4
1,2,3,4
$\rm 0.2001~g$ The copper content in the crystal is: $\rm 0.01973~mol\cdot L^{-1} \times 11.81~mL = 2.330 \times 10^{-4}~mol$

Converted to copper content in $\rm 1~g$ crystal: $\rm 2.330 \times 10^{-4} \times 1 / 0.2001 = 1.164 \times 10^{-3}~mol\cdot g^{-1}$

<Checkpoint>[$\rm 1~g$ crystal contains $\rm 1.164 \times 10^{-3}~mol~Cu$](1)</Checkpoint>

From $\rm I_2 + 2S_2O_3^{2-} = 2I^- + S_4O_6^{2-}$, then $\rm 1S_2O_3^{2-} \sim 1I^- \sim 1e^-$

The amount of electrons transferred by $\rm 1~g$ crystal is: $\rm 0.1024~mol\cdot L^{-1} \times 42.16~mL \times 1 / 0.2059 = 2.097 \times 10^{-2}~mol\cdot g^{-1}$

The number of electrons transferred per mole of crystal reacting with $\rm I^-$ is: $\rm (2.097 \times 10^{-2}) / (1.164 \times 10^{-3}) = 18.02 \approx 18$

Each mole of $\rm HIO_6^{4-}$ is reduced to $\rm I^-$ transferring $\rm 8~mol$ electrons, so each mole of crystal contains $\rm 2~mol$ ligand $\rm HIO_6^{4-}$

<Checkpoint>[Each mole of crystal contains $\rm 2~mol$ ligand $\rm HIO_6^{4-}$](1)</Checkpoint>

The remaining $\rm 2~mol$ electrons, copper is reduced to $\rm +1$ oxidation state by $\rm I^-$, so copper in the crystal is in the $\rm +3$ oxidation state, statement 1 is incorrect

<Checkpoint>[Copper in the crystal is in the $\rm +3$ oxidation state](1)</Checkpoint>

The molar mass of the crystal is: $\rm 0.2001~g / (2.3301 \times 10^{-4}~mol) = 858.8~g\cdot mol^{-1}$
The number of crystal water molecules is: $\rm 858.8~g\cdot mol^{-1} \times 25.18~\% / 18.02~g\cdot mol^{-1} \approx 12$, statement 2 is correct

<Checkpoint>[The chemical formula of this crystal contains 12 crystal water molecules](1)</Checkpoint>

Based on the inference of charge conservation, the crystal also contains a total of $\rm 5$ $\rm Na^+$ or $\rm K^+$, and their total molar mass is:
\[ 858.8 - 12 \times 18.02 - 63.55 - 2 \times (1.008 + 126.9 + 16.00 \times 6) = 131.2~(\rm g\cdot mol^{-1}) \]
Which is $\rm 1$ $\rm K^+$ and $\rm 4$ $\rm Na^+$

The chemical formula is: $\rm Na_4K[Cu(HIO_6)_2] \cdot 12H_2O$

<Checkpoint>[The chemical formula of this crystal is: $\rm Na_4K[Cu(HIO_6)_2] \cdot 12H_2O$](1)</Checkpoint>

Reaction 1 produced the corresponding potassium salt, and then $\rm NaOH$ was added to replace $\rm K^+$ to obtain $\rm Na_4K[Cu(HIO_6)_2] \cdot 12H_2O$. Here, $\rm S_2O_8^{2-}$ is reduced to $\rm SO_4^{2-}$, and $\rm H_2O$ is also produced. The equation is: $\rm 2CuSO_4 + 4H_3IO_6 + K_2S_2O_8 + 16KOH = 2K_5[Cu(HIO_6)_2] + 4K_2SO_4 + 16H_2O$, the sum of the coefficients on the left and right sides is 45, statement 3 is correct

<Checkpoint>[Reaction 1 equation is: $\rm 2CuSO_4 + 4H_3IO_6 + K_2S_2O_8 + 16KOH = 2K_5[Cu(HIO_6)_2] + 4K_2SO_4 + 16H_2O$](1)</Checkpoint>

Reaction 2 produced $\rm CuI$, $\rm I_2$, $\rm H_2O$ and sulfate. The equation is: $\rm Na_4K[Cu(HIO_6)_2] + 11H_2SO_4 + 17KI = CuI + 9I_2 + 12H_2O + 2Na_2SO_4 + 9K_2SO_4$, the sum of the coefficients on the left side is 29, the sum of the coefficients on the right side is 33, the difference is 4, statement 4 is incorrect

<Checkpoint>[Reaction 2 equation is: $\rm Na_4K[Cu(HIO_6)_2] + 11H_2SO_4 + 17KI = CuI + 9I_2 + 12H_2O + 2Na_2SO_4 + 9K_2SO_4$](1)</Checkpoint>

Statements 2 and 3 are correct, choose I
$\rm 0.2001~g$晶体中铜的含量为：$\rm 0.01973~mol\cdot L^{-1} \times 11.81~mL = 2.330 \times 10^{-4}~mol$

转化为$\rm 1~g$晶体中的铜含量：$\rm 2.330 \times 10^{-4} \times 1 / 0.2001 = 1.164 \times 10^{-3}~mol\cdot g^{-1}$

<Checkpoint>[$\rm 1~g$晶体含$\rm 1.164 \times 10^{-3}~mol~Cu$](1)</Checkpoint>

由 $\rm I_2 + 2S_2O_3^{2-} = 2I^- + S_4O_6^{2-}$，则$\rm 1S_2O_3^{2-} \sim 1I^- \sim 1e^-$

$\rm 1~g$ 晶体转移电子的量为：$\rm 0.1024~mol\cdot L^{-1} \times 42.16~mL \times 1 / 0.2059 = 2.097 \times 10^{-2}~mol\cdot g^{-1}$

每摩尔晶体与 $\rm I^-$ 反应转移的电子数为：$\rm (2.097 \times 10^{-2}) / (1.164 \times 10^{-3}) = 18.02 \approx 18$

每摩尔$\rm HIO_6^{4-}$还原到$\rm I^-$转移$\rm 8~mol$电子，故每摩尔晶体中有$\rm 2~mol$ 配体 $\rm HIO_6^{4-}$

<Checkpoint>[每摩尔晶体中有$\rm 2~mol$ 配体 $\rm HIO_6^{4-}$](1)</Checkpoint>

剩余$\rm 2~mol$ 电子，铜被$\rm I^-$还原至$\rm +1$氧化态，故晶体中铜为$\rm +3$氧化态，说法1错误

<Checkpoint>[晶体中铜为$\rm +3$氧化态](1)</Checkpoint>

晶体的摩尔质量为：$\rm 0.2001~g / (2.3301 \times 10^{-4}~mol) = 858.8~g\cdot mol^{-1}$
结晶水的数目为：$\rm 858.8~g\cdot mol^{-1} \times 25.18~\% / 18.02~g\cdot mol^{-1} \approx 12$，说法2正确

<Checkpoint>[该晶体的化学式中含12个结晶水](1)</Checkpoint>

根据电荷守恒推断，晶体中还有 $\rm Na^+$ 或 $\rm K^+$ 共 $\rm 5$ 个，其摩尔质量总计为：
\[ 858.8 - 12 \times 18.02 - 63.55 - 2 \times (1.008 + 126.9 + 16.00 \times 6) = 131.2~(\rm g\cdot mol^{-1}) \]
为 $\rm 1$ 个 $\rm K^+$ 和 $\rm 4$ 个 $\rm Na^+$

化学式为：$\rm Na_4K[Cu(HIO_6)_2] \cdot 12H_2O$

<Checkpoint>[该晶体化学式为：$\rm Na_4K[Cu(HIO_6)_2] \cdot 12H_2O$](1)</Checkpoint>

反应1产生了相应的钾盐，随后加入$\rm NaOH$替换$\rm K^+$得到$\rm Na_4K[Cu(HIO_6)_2] \cdot 12H_2O$。这里$\rm S_2O_8^{2-}$被还原为$\rm SO_4^{2-}$，此外还产生了$\rm H_2O$，方程式为：$\rm 2CuSO_4 + 4H_5IO_6 + K_2S_2O_8 + 16KOH = 2K_5[Cu(HIO_6)_2] + 4K_2SO_4 + 16H_2O$，左右两侧系数之和为45，说法3正确

<Checkpoint>[反应1方程式为：$\rm 2CuSO_4 + 4H_5IO_6 + K_2S_2O_8 + 16KOH = 2K_5[Cu(HIO_6)_2] + 4K_2SO_4 + 16H_2O$](1)</Checkpoint>

反应2中产生了$\rm CuI$、$\rm I_2$、$\rm H_2O$以及硫酸盐，方程式为：$\rm Na_4K[Cu(HIO_6)_2] + 11H_2SO_4 + 17KI = CuI + 9I_2 + 12H_2O + 2Na_2SO_4 + 9K_2SO_4$，左侧系数之和为29，右侧系数之和为33，差为4，说法4错误

<Checkpoint>[反应2方程式为：$\rm Na_4K[Cu(HIO_6)_2] + 11H_2SO_4 + 17KI = CuI + 9I_2 + 12H_2O + 2Na_2SO_4 + 9K_2SO_4$](1)</Checkpoint>

说法2和3正确，选I