Thermogravimetric experiments were performed on the synthesized potassium tris(oxalate)ferrate(III) crystals. The thermogravimetric curve in the range of $0-650\degree\mathrm{C}$ showed four weight loss peaks: a weight loss of $10.71\%$ at $99\degree\mathrm{C}$, a weight loss of $8.90\%$ at $287\degree\mathrm{C}$, a weight loss of $14.85\%$ at $435\degree\mathrm{C}$, and a weight loss of $8.83\%$ at $450-580\degree\mathrm{C}$.

Due to the systematic errors present in thermogravimetric experiments, there may be some differences between the theoretical and experimental weight losses. Based on the thermogravimetric data, the four-step chemical equations for the thermal decomposition of potassium tris(oxalate)ferrate(III) crystals can be written (with the simplest integer ratio coefficients). Regarding these four weight loss peaks and the corresponding decomposition chemical equations, the following options are correct:
对实验合成出的三草酸合铁(III)酸钾晶体进行热重实验，在$0-650\degree\mathrm{C}$的热重曲线存在四个失重峰：在$99\degree\mathrm{C}$失重$10.71\%$，在$287\degree\mathrm{C}$失重$8.90\%$，在$435\degree\mathrm{C}$失重$14.85\%$，在$450-580\degree\mathrm{C}$失重$8.83\%$。

由于热重实验存在的系统误差，理论失重和实验失重可能存在些许差异。通过热重数据，可以写出对应的三草酸合铁(III)酸钾晶体受热分解的四步化学方程式（系数为最简整数比）。关于这四个失重峰和对应的分解化学方程式，下列选项正确的是
The difference between the theoretical weight loss value and the experimental weight loss value of the first weight loss peak is $0.39\%$.
The decomposition chemical equation corresponding to the second weight loss peak has two gaseous products.
The sum of the stoichiometric coefficients of the products in the decomposition chemical equation corresponding to the second weight loss peak is an odd number.
The decomposition chemical equation corresponding to the third weight loss peak has only one gaseous product.
The decomposition chemical equation corresponding to the third weight loss peak has two kinds of oxide products.
The theoretical weight loss value of the fourth weight loss peak is less than the experimental weight loss value.
The chemical formula of the final remaining solid product in the fourth weight loss peak contains five atoms.
第一个失重峰的理论失重值与实验失重值差距为$0.39\%$。
第二个失重峰对应的分解化学方程式有两种气体产物
第二个失重峰对应的分解化学方程式产物系数和为奇数。
第三个失重峰对应的分解化学方程式只有一种气体产物。
第三个失重峰对应的分解化学方程式产物有两种氧化物。
第四个失重峰的理论失重值小于实验失重值。
第四个失重峰的最后剩余的固体产物化学式含有五个原子。
The chemical formula of potassium ferrioxalate is $\mathrm{K_3[Fe(C_2O_4)_3]}$, but because it is an experimentally synthesized crystal, it contains water of crystallization, and the crystal chemical formula can be $\mathrm{K_3[Fe(C_2O_4)_3]\cdot xH_2O}$, whose formula weight is $437.21+18.016x$

<Checkpoint>[The crystal chemical formula can be $\mathrm{K_3[Fe(C_2O_4)_3]\cdot xH_2O}$, whose formula weight is $437.21+18.016x$](1)</Checkpoint>

The first step of thermogravimetric analysis loses $10.71\%$. The temperature is low, considering it is a decomposition reaction that loses water of crystallization.

<Checkpoint>[The temperature of the first step of thermogravimetric analysis is low, which is a decomposition reaction that loses water of crystallization](1)</Checkpoint>

Assuming that y water molecules are lost, the first step satisfies the equation

$\frac{18.016y}{437.21 + 18.016x} = 0.1071$

Substituting the values of y to solve for x, we can get $y = 3$, $x = 3.74$, and $y = 4$, $x = 13.08$. Therefore, $y = 3$ is a more reasonable solution. Therefore, three water molecules of crystallization are lost in the first step, and the remaining water of crystallization may be 0 or 1: the theoretical weight loss in the two cases is $11.00\%$ and $10.61\%$ respectively. Now let's consider the second reaction. When there are 0 remaining water molecules of crystallization, the molecular weight lost in the second step of decomposition is $\mathrm{491.26\times0.0890=43.72}$, which is close to the molecular weight of carbon dioxide. When there is 1 remaining water molecule of crystallization, the molecular weight lost in the second step of the reaction is $\mathrm{509.27\times0.0890=45.33}$. After removing one water molecule, the remaining is 27.31, which is close to the molecular weight of carbon monoxide. However, the oxidation number of carbon in oxalate is +3, and that in carbon monoxide is +2; therefore, the oxalate is reduced in this process, and no reasonable reducing agent can be found in the system at this time, so this process does not conform to the actual situation. Therefore, the first step of the decomposition reaction is the loss of 3 water molecules, and at the same time, 0 water molecules of crystallization remain.

<Checkpoint>[The first step of the reaction loses 3 water molecules, and at the same time, 0 water molecules remain](1)</Checkpoint>

The corresponding equation is

$\mathrm{K_3[Fe(C_2O_4)_3]\cdot 3H_2O=K_3[Fe(C_2O_4)_3]+3H_2O}$

<Checkpoint>[$\mathrm{K_3[Fe(C_2O_4)_3]\cdot 3H_2O=K_3[Fe(C_2O_4)_3]+3H_2O}$](1)</Checkpoint>

The theoretical weight loss value of the first step is $\mathrm{18.016\times3/(437.21+18.016\times3)\times100\%=11.00\%}$, and the difference from the experiment is $0.29\%$, so option A is wrong.

<Checkpoint>[The difference between the theoretical weight loss value of the first step and the experiment is $0.29\%$, so option A is wrong](1)</Checkpoint>

With the molecular weight of the substrate, the molecular weight lost in the second step of the decomposition reaction is $\mathrm{491.26\times0.0890=43.72}$, which is close to the molecular weight of carbon dioxide 44.01. Considering the thermogravimetric error, the second step of the decomposition reaction is the reaction of losing one molecule of carbon dioxide, so there is only one gaseous product, and option B is wrong.

<Checkpoint>[The second step of the decomposition reaction loses one molecule of carbon dioxide, so there is only one gaseous product, and option B is wrong](1)</Checkpoint>

At this time, the only oxidizing agent can be trivalent iron in the molecule, so the decomposition equation can be written as:

$\mathrm{2K_3[Fe(C_2O_4)_3]=2CO_2+2K_2[Fe(C_2O_4)_2]+K_2C_2O_4}$

The decomposition equation can also be written as

$\mathrm{2K_3[Fe(C_2O_4)_3]=2CO_2+2FeC_2O_4+3K_2C_2O_4}$

<Checkpoint>[$\mathrm{2K_3[Fe(C_2O_4)_3]=2CO_2+2K_2[Fe(C_2O_4)_2]+K_2C_2O_4}$ or $\mathrm{2K_3[Fe(C_2O_4)_3]=2CO_2+2FeC_2O_4+3K_2C_2O_4}$](1)</Checkpoint>

The sum of product coefficients is 5 or 7, which are odd numbers, so option C is correct.

<Checkpoint>[The sum of product coefficients is 5 or 7, which is an odd number, so option C is correct.](0.5)</Checkpoint>

The molecular weight lost in the third step of the decomposition reaction is $\mathrm{491.26\times0.1485=72.95}$. Since the gaseous product can only be a carbon oxide, the decomposition product can be set as $\mathrm{C_xO_y}$. It can be known by trial that when it is $\mathrm{C_2O_3=CO_2+CO}$, the molecular weight is 72.02, which is close to 72.95. Therefore, the third step of the decomposition reaction produces one equivalent of $\mathrm{CO_2+CO}$, two gases, and option D is wrong.

<Checkpoint>[The third step of the decomposition reaction produces one equivalent of $\mathrm{CO_2+CO}$](1)</Checkpoint>

<Checkpoint>[Two gases, option D is wrong](0.5)</Checkpoint>

Since there is no trivalent iron in the system at this time, it is a disproportionation reaction of the carbon atom itself. One equivalent of $\mathrm{CO_2+CO}$ indicates that only one oxalate is decomposed. Considering the stability of ferrous oxalate and potassium oxalate, the stronger the polarizing power of the cation, the weaker the stability of the oxoacid radical; ferrous ions, as transition metals, have significantly stronger polarizing power than alkali metal potassium ions, so ferrous oxalate decomposes first.

<Checkpoint>[The polarizing power of ferrous ions is stronger, and the stability of ferrous oxalate is lower than that of potassium oxalate](1)</Checkpoint>

Therefore, the decomposition equation is written as:

$\mathrm{K_2[Fe(C_2O_4)_2]=FeO+CO+CO_2+K_2C_2O_4}$

<Checkpoint>[$\mathrm{K_2[Fe(C_2O_4)_2]=FeO+CO+CO_2+K_2C_2O_4}$](1)</Checkpoint>

There are three kinds of oxides in the product, so option E is wrong.

<Checkpoint>[There are three kinds of oxides in the product, so option E is wrong.](0.5)</Checkpoint>

The molecular weight lost in the fourth step of decomposition is
$\mathrm{491.26\times0.0883=43.37}$, corresponding to 1.5 times the molecular weight of carbon monoxide. At this time, the carbon-containing species in the system are only two equivalents of potassium oxalate, so the reaction is the decomposition of potassium oxalate into potassium carbonate:

<Checkpoint>[In the fourth step of weight loss, two molecules of carbon monoxide are produced, and potassium oxalate is decomposed into potassium carbonate](1)</Checkpoint>

$\mathrm{K_2C_2O_4=K_2CO_3+CO}$

<Checkpoint>[$\mathrm{K_2C_2O_4=K_2CO_3+CO}$](1)</Checkpoint>

The theoretical weight loss value is $\mathrm{44.01/491.26=0.0896}$, which is greater than the experimental weight loss value, so option F is wrong.

<Checkpoint>[The theoretical weight loss value is $\mathrm{8.96\%}$, which is greater than the experimental weight loss value, so option F is wrong.](1)</Checkpoint>

The product potassium carbonate is a compound containing six atoms, so option G is wrong.

<Checkpoint>[The product potassium carbonate is a compound containing six atoms, so option G is wrong.](0.5)</Checkpoint>

In summary, option C is correct.

<Checkpoint>[In summary, option C is correct.](0.5)</Checkpoint>
三草酸合铁(III)酸钾的化学式为$\mathrm{K_3[Fe(C_2O_4)_3]}$，但因为是实验合成的晶体，因而存在结晶水，晶体化学式可为$\mathrm{K_3[Fe(C_2O_4)_3]\cdot xH_2O}$，其式量为$437.21+18.016x$

<Checkpoint>[晶体化学式可为$\mathrm{K_3[Fe(C_2O_4)_3]\cdot xH_2O}$，其式量为$437.21+18.016x$](1)</Checkpoint>

第一步热重失去$10.71\%$，温度较低，考虑是失去结晶水的分解反应。

<Checkpoint>[第一步热重温度较低，是失去结晶水的分解反应](1)</Checkpoint>

假设失去了y个水，则第一步的过程满足方程式

$\frac{18.016y}{437.21 + 18.016x} = 0.1071$

分别代入y的数值求解x，可以得到$y = 3$时，$x = 3.74$, 而$y = 4$时，$x = 13.08$，因此$y = 3$是比较合理的解。故第一步反应中失去了三个结晶水，而剩余的结晶水可能是0个或者1个：两种情况下理论失重分别为$11.00%$和$10.61%$。下面我们先考虑第二个反应。当剩余0个结晶水时，第二步分解失去的分子量为$\mathrm{491.26\times0.0890=43.72}$，接近二氧化碳的分子量。当剩余1个结晶水时，第二步反应失去的分子量为$\mathrm{509.27\times0.0890=45.33}$，去掉一分子水，剩余27.31，接近一氧化碳的分子量。然而草酸根中碳为+3氧化数，一氧化碳中为+2氧化数；因此在该过程中草酸根被还原，而此时体系中找不到合理的还原剂，故该过程不符合实际情况。因此第一步分解反应为失去3个水，同时剩余0个结晶水。

<Checkpoint>[第一步反应失去3分子水，同时剩余0分子水](1)</Checkpoint>

相应的方程式为

$\mathrm{K_3[Fe(C_2O_4)_3]\cdot 3H_2O=K_3[Fe(C_2O_4)_3]+3H_2O}$

<Checkpoint>[$\mathrm{K_3[Fe(C_2O_4)_3]\cdot 3H_2O=K_3[Fe(C_2O_4)_3]+3H_2O}$](1)</Checkpoint>

第一步的理论失重值为$\mathrm{18.016\times3/(437.21+18.016\times3)\times100\%=11.00\%}$，与实验差距为$0.29\%$，选项A错误。

<Checkpoint>[第一步的理论失重值与实验差距为$0.29\%$，选项A错误](1)</Checkpoint>

有了底物分子量，第二步分解反应失去的分子量为$\mathrm{491.26\times0.0890=43.72}$，接近二氧化碳的分子量44.01。考虑到热重误差，第二步分解反应就是失去一分子二氧化碳的反应，从而只有一种气体产物，选项B错误。

<Checkpoint>[第二步分解反应失去一分子二氧化碳，从而只有一种气体产物，选项B错误](1)</Checkpoint>

此时，充当氧化剂的只能为分子内的三价铁，从而可以写出分解方程式：

$\mathrm{2K_3[Fe(C_2O_4)_3]=2CO_2+2K_2[Fe(C_2O_4)_2]+K_2C_2O_4}$

分解方程式也可以写为

$\mathrm{2K_3[Fe(C_2O_4)_3]=2CO_2+2FeC_2O_4+3K_2C_2O_4}$

<Checkpoint>[$\mathrm{2K_3[Fe(C_2O_4)_3]=2CO_2+2K_2[Fe(C_2O_4)_2]+K_2C_2O_4}$ 或 $\mathrm{2K_3[Fe(C_2O_4)_3]=2CO_2+2FeC_2O_4+3K_2C_2O_4}$](1)</Checkpoint>

产物系数和为5或7，均为奇数，选项C正确。

<Checkpoint>[产物系数和为5或7，为奇数，选项C正确。](0.5)</Checkpoint>

第三步分解反应失去的分子量为$\mathrm{491.26\times0.1485=72.95}$，由于气体产物只能为碳氧化合物，可以设分解产物为$\mathrm{C_xO_y}$，经尝试可知为$\mathrm{C_2O_3=CO_2+CO}$时，分子量为72.02，与72.95接近，从而第三步分解反应产生一当量$\mathrm{CO_2+CO}$，两种气体，选项D错误。

<Checkpoint>[第三步分解反应产生一当量$\mathrm{CO_2+CO}$](1)</Checkpoint>

<Checkpoint>[两种气体，选项D错误](0.5)</Checkpoint>

由于此时体系不存在三价铁，所以是碳原子自身的歧化反应，一当量$\mathrm{CO_2+CO}$表明只有一分子草酸根被分解。考虑草酸亚铁和草酸钾的稳定性，阳离子的极化能力越强，含氧酸根的稳定性越弱；亚铁离子作为过渡金属，极化能力明显要强于碱金属钾离子，故草酸亚铁先发生分解。

<Checkpoint>[亚铁离子的极化能力更强，草酸亚铁的稳定性低于草酸钾](1)</Checkpoint>

因此写出分解方程式：

$\mathrm{K_2[Fe(C_2O_4)_2]=FeO+CO+CO_2+K_2C_2O_4}$

<Checkpoint>[$\mathrm{K_2[Fe(C_2O_4)_2]=FeO+CO+CO_2+K_2C_2O_4}$](1)</Checkpoint>

产物有三种氧化物，选项E错误。

<Checkpoint>[产物有三种氧化物，选项E错误。](0.5)</Checkpoint>

第四步分解，失去的分子量为
$\mathrm{491.26\times0.0883=43.37}$，对应1.5个一氧化碳的分子量。此时体系含碳物种仅为两当量的草酸钾，因此该反应为草酸钾分解为碳酸钾：

<Checkpoint>[第四步失重，产生两分子一氧化碳，草酸钾分解为碳酸钾](1)</Checkpoint>

$\mathrm{K_2C_2O_4=K_2CO_3+CO}$

<Checkpoint>[$\mathrm{K_2C_2O_4=K_2CO_3+CO}$](1)</Checkpoint>

理论失重值为$\mathrm{44.01/491.26=0.0896}$，大于实验失重值，选项F错误。

<Checkpoint>[理论失重值为$\mathrm{8.96\%}$，大于实验失重值，选项F错误。](1)</Checkpoint>

产物碳酸钾为含六个原子的化合物，选项G错误。

<Checkpoint>[产物碳酸钾为含六个原子的化合物，选项G错误。](0.5)</Checkpoint>

综上，选项C正确。

<Checkpoint>[综上，选项C正确。](0.5)</Checkpoint>