Dissolve $\rm MnCl_2\cdot 4H_2O$ in ethanol, and add an aqueous solution of $\rm Htopca$ (a carboxylic acid derivative with the molecular formula $\rm C_5H_6N_2O_3$). After mixing and stirring at 80 ℃ for 2 h, filter. The filtrate is slowly evaporated at room temperature, and a yellow blocky crystal precipitates from the solution. Thermogravimetric analysis of $\bf X$ under a nitrogen atmosphere shows that $\bf X$ loses a% of its weight between 65 and 120 ℃, corresponding to the loss of all crystal water and structural water; continue heating to 250 ℃, the weight loss is b%; continue heating to 850 ℃, the weight loss is further 14.04%, and the subsequent solid mass no longer changes. Chemical analysis of the residual solid shows that the solid contains two substances, and the mass fraction of $\rm Mn$ in the mixture is 46.17%. The mixed gas released during the second weight loss is analyzed, and the mixed gas contains three gas molecules, of which only propyne contains carbon. After cooling the mixed gas, it is fully absorbed by bromine water. The mass of the remaining gas is 37.94% of the original gas mass, and the remaining gas changes from colorless to reddish-brown in the air. It is known that $\bf X$ is a mononuclear $\rm Mn$ complex, and the oxidation state of $\rm Mn$ does not change during the thermogravimetric analysis.

Given a+b=58.07, which of the following statements are correct:
1. The sum of the number of all atoms in the chemical formula of $\bf X$ is less than 45
2. The three gases produced by the second weight loss are in a ratio of 1:1:2
3. In the total reaction equation for the second and third steps of weight loss, the sum of the coefficients on the right side is less than 12
4. The final product of the weight loss can react with concentrated sulfuric acid, producing only one gas
将$\rm MnCl_2\cdot 4H_2O$溶于乙醇中，并加入$\rm Htopca$（一种羧酸衍生物，分子式为$\rm C_5H_6N_2O_3$）的水溶液，在80 ℃下混合搅拌2 h后过滤，滤液在室温下缓慢蒸发，溶液中析出一种黄色块状晶体。对$\bf X$在氮气气氛下进行热重分析实验表明：在65~120 ℃下$\bf X$失重a%，对应于失去所有结晶水和结构水；继续加热至250 ℃，失重b%；继续加热至850 ℃，再失重14.04%，后续固体质量不再发生变化。对残留固体化学分析表明：固体中含有两种物质，混合物中$\rm Mn$的质量分数为46.17%。对第二次失重放出的混合气体进行分析，混合气体中含有三种气体分子，其中含碳元素的气体仅有丙炔；将该混合气体冷却后用溴水充分吸收，剩余气体质量为原气体质量的37.94%，且剩余气体在空气中会由无色变为红棕色。已知$\bf X$为单核$\rm Mn$配合物，在热重分析的过程中$\rm Mn$没有发生氧化态的改变。

已知a+b=58.07，下列说法中正确的有：
1. $\bf X$化学式中所有原子数目之和小于45
2. 第二次失重产生的三种气体，比例为1:1:2
3. 第二步和第三步失重的总反应方程式中，右侧系数之和小于12
4. 失重的最终产物可以与浓硫酸反应，且只产生一种气体
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
Since the remaining gas changes from colorless to reddish-brown in air, and considering that $\rm Htopca$ contains nitrogen, it is speculated that the remaining gas contains $\rm NO$; since the gas first underwent cooling, it is speculated that it may also contain $\rm H_2O$ before cooling.

Therefore, the composition of the mixed gas is likely propyne + $\rm NO$ + $\rm H_2O$, or propyne + $\rm NO$ + another gas.

<Checkpoint>[The composition of the mixed gas is likely propyne + $\rm NO$ + $\rm H_2O$, or propyne + $\rm NO$ + another gas.](1)</Checkpoint>

Since the mass of the remaining gas is 37.94% of the original gas, assuming it is all $\rm NO$, then:

$\rm Mr(remaining\ 1) = (14.01 + 16.00)/0.3794 - (14.01 + 16.00) = 49.09$

Subtracting the relative molecular mass of one molecule of propyne yields:

$\rm Mr(remaining\ 2) = 49.09 - 40.06 = 9.03$

This corresponds exactly to 0.5 $\rm H_2O$, which is consistent with the prediction.
Therefore, the gas lost in the second weight loss is $\rm 2C_3H_4 + 2NO + H_2O$, so statement 2 is incorrect.

<Checkpoint>[The ratio of the gas lost in the second weight loss is $\rm C_3H_4:NO:H_2O=2:2:1$](2)</Checkpoint>

Since $\rm Mn$ does not change during the process, the $\rm Mn$ in the residual solid should be an oxide of the corresponding valence state, namely $\rm MnO$. Combined with the mass fraction, we have $\rm Mr(remaining) = 54.94 / 0.4617 - 54.94 - 16 = 48.06$, which is found to be exactly the relative mass of 4 carbons, so the molecular composition of the residual solid is $\rm MnO + 4C$. The reaction of elemental carbon with concentrated sulfuric acid can produce $\rm CO_2$ and $\rm SO_2$, so statement 4 is incorrect.

<Checkpoint>[The residual solid is $\rm MnO:C=1:4$](1)</Checkpoint>

Therefore, the molecular weight of $\bf X$ is $\rm Mr = 119.00 / (1 - 0.5807 - 0.1404) = 426.66$

The first and second weight losses $\rm M = 426.66 \times 0.5807 = 247.76$. Removing $\rm 2C_3H_4 + 2NO + H_2O$ leaves approximately 90, so the first loss is $\rm 5H_2O$

The third weight loss $\rm M = 426.66 \times 0.1404 = 59.90$, corresponding to $\rm 2NO$

<Checkpoint>[The third loss is $\rm NO$](1)</Checkpoint>

Therefore, the molecular formula of $\bf X$ is $\rm MnC_{10}H_{10}N_4O_6\cdot 5H_2O$, i.e., $\rm Mn(topca)_2(H_2O)_5$

<Checkpoint>[The molecular formula of $\bf X$ is $\rm MnC_{10}H_{10}N_4O_6\cdot 5H_2O$, i.e., $\rm Mn(topca)_2(H_2O)_5$](2)</Checkpoint>

The sum of the number of all atoms in the chemical formula of $\bf X$ is 46, so statement 1 is incorrect.

The second and third steps are the loss of $\rm 2C_3H_4 + 4NO + H_2O$ from 1 $\bf X$, producing $\rm MnO + 4C$. The equation is $\rm Mn(topca)_2 \to 2C_3H_4 + 4NO + H_2O + MnO + 4C$, and the sum of the coefficients on the right side is 12, so statement 3 is incorrect.

<Checkpoint>[The total equation for the second and third weight loss steps is $\rm Mn(topca)_2 \to 2C_3H_4 + 4NO + H_2O + MnO + 4C$](1)</Checkpoint>

Choose A
由于剩余气体在空气中会由无色变为红棕色，结合$\rm Htopca$中含有氮元素，推测剩余气体中包含$\rm NO$；由于气体首先经历了冷却，因此推测冷却前还可能包含$\rm H_2O$。

因此混合气体的组成很可能是丙炔+$\rm NO$+$\rm H_2O$，或者是丙炔+$\rm NO$+一种其他气体。

<Checkpoint>[混合气体的组成很可能是丙炔+$\rm NO$+$\rm H_2O$，或者是丙炔+$\rm NO$+一种其他气体。](1)</Checkpoint>

由于剩余气体质量为原气体的37.94%，假设全部为$\rm NO$，则有：

$\rm Mr(剩余1) = (14.01 + 16.00)/0.3794 - (14.01 + 16.00) = 49.09$

再减去一分子丙炔的相对分子质量后可得：

$\rm Mr(剩余2) = 49.09 - 40.06 = 9.03$

恰好对应于0.5个$\rm H_2O$，符合预测
故第二次失重气体为$\rm 2C_3H_4 + 2NO + H_2O$，说法2错误

<Checkpoint>[第二次失重气体比例为$\rm C_3H_4:NO:H_2O=2:2:1$](2)</Checkpoint>

由于$\rm Mn$在过程中没有发生变化，因此残余固体中$\rm Mn$应为对应价态的氧化物，即$\rm MnO$。结合质量分数，有$\rm Mr(剩余) = 54.94 / 0.4617 - 54.94 - 16 = 48.06$，发现恰好为4个碳的相对质量，故残余固体的分子组成为$\rm MnO + 4C$。碳单质与浓硫酸反应可以产生$\rm CO_2$和$\rm SO_2$，说法4错误

<Checkpoint>[残余固体为$\rm MnO:C=1:4$](1)</Checkpoint>

故$\bf X$的分子量$\rm Mr = 119.00 / (1 - 0.5807 - 0.1404) = 426.66$

第一、二次失重$\rm M = 426.66 \times 0.5807 = 247.76$除去$\rm 2C_3H_4 + 2NO + H_2O$剩下大约90，因此第一次失去$\rm 5H_2O$

第三次失重$\rm M = 426.66 \times 0.1404 = 59.90$，对应于$\rm 2NO$

<Checkpoint>[第三次失去$\rm NO$](1)</Checkpoint>

因此$\bf X$的分子式为$\rm MnC_{10}H_{10}N_4O_6\cdot 5H_2O$，即$\rm Mn(topca)_2(H_2O)_5$

<Checkpoint>[$\bf X$的分子式为$\rm MnC_{10}H_{10}N_4O_6\cdot 5H_2O$，即$\rm Mn(topca)_2(H_2O)_5$](2)</Checkpoint>

$\bf X$化学式中所有原子数目之和为46，说法1错误

第二、三步为1个$\bf X$失去$\rm 2C_3H_4 + 4NO + H_2O$，产生$\rm MnO + 4C$，方程式为$\rm Mn(topca)_2 \to 2C_3H_4 + 4NO + H_2O + MnO + 4C$，右侧系数和为12，说法3错误

<Checkpoint>[第二、三步失重总方程式为$\rm Mn(topca)_2 \to 2C_3H_4 + 4NO + H_2O + MnO + 4C$](1)</Checkpoint>

选A