$\mathrm{N}_{2} \mathrm{O}$ emits a popping sound when it reacts with $\mathrm{CS}_{2}$. To investigate the products of this reaction, an interest group conducted the following experiment:

A dry and clean test tube was weighed, and its weight was $31.33 \mathrm{~g}$. It was then filled with $\mathrm{N}_{2} \mathrm{O}$ under normal pressure. A few drops of $\mathrm{CS}_{2}$ were added. After the $\mathrm{N}_{2} \mathrm{O}$ had reacted almost completely, the generated gas was passed sequentially through a. saturated $\mathrm{Na}_{2} \mathrm{CO}_{3}$ solution, concentrated sulfuric acid; b. $\mathrm{I}_{2} \mathrm{O}_{5}$ solid; c. saturated $\mathrm{Na}_{2} \mathrm{CO}_{3}$ solution, concentrated sulfuric acid. Finally, the volume of the collected gas was similar to the volume of the initial $\mathrm{N_2O}$ gas in the test tube. It is known that the mass increase in a. is approximately 4 times the mass decrease in b. After the reaction is complete, the unreacted $\mathrm{CS}_{2}$ in the test tube is removed and weighed, and the weight is $31.70 \mathrm{~g}$.

It is known that there are 4 kinds of molecules in the products. According to the above information, write the equation for the reaction, paying attention to the actual existing forms of each substance's molecules. Regarding the product of the coefficients in the equation, the correct option is:
$\mathrm{N}_{2} \mathrm{O}$ 与 $\mathrm{CS}_{2}$ 作用时会发出爆鸣声，为了探究这个反应的产物,某兴趣小组做了如下实验:

对一干燥洁净的试管进行称重, 其重量为 $31.33 \mathrm{~g}$, 现将其充满常压下的 $\mathrm{N}_{2} \mathrm{O}$, 加入几滴$\mathrm{CS}_{2}$, 待 $\mathrm{N}_{2} \mathrm{O}$ 基本完全反应后, 将产生的气体先后通入a. 饱和$\mathrm{Na} \mathrm{CO}_{3}$ 溶液、浓硫酸; b. $\mathrm{I}_{2} \mathrm{O}_{5}$ 固体; c. 饱和 $\mathrm{Na}_{2} \mathrm{CO}_{3}$ 溶液、浓硫酸。最后收集到气体与最开始试管中$\mathrm{N_2O}$气体体积相近, 已知a.增重的质量大概为b.
减重质量的 4 倍, 反应完毕后, 除去试管中未反应的 $\mathrm{CS}_{2}$ 并称重，重量为 $31.70 \mathrm{~g}$。

已知产物中有 4 种分子, 根据以上信息, 写出反应的方程式，注意各物质分子的实际存在形式。关于方程式各系数乘积，选项正确的为：
All other options are incorrect
294912
9
8
521420800
11855241216
300
16384
其他选项均不正确
294912
9
8
521420800
11855241216
300
16384
First, based on the reactants and experimental phenomena, speculate on the possible products:
The reaction between $\mathrm{N_2O}$ and $\mathrm{CS_2}$ is a typical redox reaction. The nitrogen element in $\mathrm{N_2O}$ has oxidizing properties, while the sulfur element in $\mathrm{CS_2}$ has reducing properties, possibly producing $\mathrm{N_2}$ and elemental sulfur. The generated gas is passed into a. saturated $\mathrm{Na_2CO_3}$ and concentrated sulfuric acid solution, indicating the generation of acidic gas, possibly $\mathrm{SO_2}$. During the reaction, $\mathrm{SO_2}$ is completely absorbed, and the weight increase corresponds to the mass of $\mathrm{SO_2}$ in the gas: $\mathrm{m(SO_2)=\Delta m(a)}$.

<Checkpoint>[The weight increase in a. $\mathrm{Na_2CO_3}$ indicates the absorption of $\mathrm{SO_2}$ in the products](1)</Checkpoint>

<Checkpoint>[The weight increase in a. corresponds to the mass of $\mathrm{SO_2}$ in the gas](1)</Checkpoint>

Passed into b. $\mathrm{I_2O_5}$ solid, causing its mass to decrease, indicating that the generated gas also contains reducing gas, and oxygen atoms are transferred to the reducing gas during the reaction with $\mathrm{I_2O_5}$. Considering that $\mathrm{SO_2}$ has been removed by $\mathrm{Na_2CO_3}$, the reducing gas may be $\mathrm{CO}$, and the reaction equation is:

$\mathrm{I_2O_5+5CO}\rightarrow \mathrm{I_2+5CO_2}$

The weight loss is the decrease of oxygen atoms, and the relationship between the mass of $\mathrm{CO}$ in the gas and the weight loss is:

$\mathrm{m(CO)=28.01 \times \frac{\Delta m(b)}{16.00}}$

<Checkpoint>[The weight loss in b. indicates that $\mathrm{I_2O_5}$ reacts with $\mathrm{CO}$ in the product gas](1)</Checkpoint>

<Checkpoint>[The weight loss in b. can be used to calculate the mass of $\mathrm{CO}$ in the gas](1)</Checkpoint>

The gas continues to be passed into c. $\mathrm{Na_2CO_3}$ and concentrated sulfuric acid solution, which should remove the gas generated in b., namely $\mathrm{CO_2}$.

<Checkpoint>[Passing into c. can remove the $\mathrm{CO_2}$ generated in b.](1)</Checkpoint>

The final unreacted gas is similar in volume to $\mathrm{N_2O}$, indicating that the gas product contains inert gas, corresponding to $\mathrm{N_2}$. The volume before and after is the same, indicating that the nitrogen atoms in $\mathrm{N_2O}$ are completely converted to $\mathrm{N_2}$.

<Checkpoint>[The gas volume is the same before and after the reaction, indicating that all $\mathrm{N_2O}$ is converted to $\mathrm{N_2}$](1)</Checkpoint>

The mass of the test tube increases after the reaction, indicating that elemental sulfur is formed. The question indicates that there are four types of molecules in the product, which is just right.

<Checkpoint>[The test tube gains weight after the reaction, indicating the formation of elemental sulfur](1)</Checkpoint>

The unbalanced reaction equation can be written first:

$\mathrm{xN_2O+zCS_2}\rightarrow\mathrm{xN_2+ySO_2+zCO+nS_8}$

The weight increase in a. is approximately 4 times the weight loss in b.: $\mathrm{m(CO)=28.01 \times \frac{\Delta m(b)}{16.00}=28.01 \times \frac{\Delta m(a)}{4\times 16.00}=0.4377\Delta m(a)=0.4377m(SO_2)}$

$\mathrm{n(CO)=m(CO)/M(CO)=0.4377m(SO_2)/M(CO)=0.4377n(SO_2)\times M(SO_2)/M(CO)=1.001 \times n(SO_2)}$

That is, the number of moles of $\mathrm{SO_2}$ and $\mathrm{CO}$ in the product are the same, $\mathrm{y=z}$.

<Checkpoint>[According to the weight relationship between a. and b., it is concluded that the number of moles of $\mathrm{SO_2}$ and $\mathrm{CO}$ are the same](1)</Checkpoint>

In fact, the equation can be solved based on the conservation of the number of atoms:
$$
\mathrm{x=2y+z}
$$
$$
\mathrm{y=z}
$$
$$
\mathrm{2z=y+8n}
$$

Reduced to the simplest form, the equation is:
$$
24 \mathrm{~N}_{2} \mathrm{O}+8 \mathrm{CS}_{2} \rightarrow 24 \mathrm{~N}_{2}+8 \mathrm{SO}_{2}+8 \mathrm{CO}+\mathrm{S}_{8}
$$

<Checkpoint>[The equation solved according to the conservation of the number of atoms is $ 24 \mathrm{~N}_{2} \mathrm{O}+8 \mathrm{CS}_{2} \rightarrow 24 \mathrm{~N}_{2}+8 \mathrm{SO}_{2}+8 \mathrm{CO}+\mathrm{S}_{8} $](1)</Checkpoint>

The product of the coefficients is 294912, so option B is correct.
首先根据反应物和实验现象，推测可能生成的物质：
$\mathrm{N_2O}$与$\mathrm{CS_2}$反应为典型的氧化还原反应，其中$\mathrm{N_2O}$氮元素具有氧化性而$\mathrm{CS_2}$硫元素具有还原性，可能生成$\mathrm{N_2}$和单质硫。生成的气体通入a. 饱和$\mathrm{Na_2CO_3}$和浓硫酸溶液发生反应，说明生成了酸性气体，可能是$\mathrm{SO_2}$，反应时$\mathrm{SO_2}$被完全吸收，增重即对应气体中$\mathrm{SO_2}$的质量：$\mathrm{m(SO_2)=\Delta m(a)}$。

<Checkpoint>[a.中$\mathrm{Na_2CO_3}$增重，说明吸收了产物中的$\mathrm{SO_2}$](1)</Checkpoint>

<Checkpoint>[a.中增重对应气体中$\mathrm{SO_2}$的质量](1)</Checkpoint>

通入b. $\mathrm{I_2O_5}$固体使其质量减小，说明生成的气体还有还原性气体，且在与$\mathrm{I_2O_5}$反应时有氧原子转移到了该还原性气体中。考虑到$\mathrm{SO_2}$已经被$\mathrm{Na_2CO_3}$除去，该还原性气体可能是$\mathrm{CO}$，反应方程式为：

$\mathrm{I_2O_5+5CO}\rightarrow \mathrm{I_2+5CO_2}$

减重为氧原子的减少，气体中$\mathrm{CO}$的质量与减重关系为：

$\mathrm{m(CO)=28.01 \times \frac{\Delta m(b)}{16.00}}$

<Checkpoint>[b.减重说明$\mathrm{I_2O_5}$与产物气体中的$\mathrm{CO}$反应](1)</Checkpoint>

<Checkpoint>[b.减重可以算出气体中$\mathrm{CO}$的质量](1)</Checkpoint>

气体继续通入c. $\mathrm{Na_2CO_3}$和浓硫酸溶液，应该是除去b.生成的气体，即$\mathrm{CO_2}$。

<Checkpoint>[通入c.可以除去b.中生成的$\mathrm{CO_2}$](1)</Checkpoint>

最终得到未反应的气体与$\mathrm{N_2O}$体积相近，说明气体产物中含有惰性气体，对应$\mathrm{N_2}$。前后体积相同，说明$\mathrm{N_2O}$中氮原子完全转化$\mathrm{N_2}$。

<Checkpoint>[反应前和处理后气体体积相同，说明$\mathrm{N_2O}$全部转化为$\mathrm{N_2}$](1)</Checkpoint>

试管反应后质量增加，说明有单质硫形成。题中给出产物有四种分子，恰好。

<Checkpoint>[反应后试管增重，说明有硫单质形成](1)</Checkpoint>

可以先写出未配平的反应方程式：

$\mathrm{xN_2O+zCS_2}\rightarrow\mathrm{xN_2+ySO_2+zCO+nS_8}$

a.增重约为b.减重的4倍：$\mathrm{m(CO)=28.01 \times \frac{\Delta m(b)}{16.00}=28.01 \times \frac{\Delta m(a)}{4\times 16.00}=0.4377\Delta m(a)=0.4377m(SO_2)}$

$\mathrm{n(CO)=m(CO)/M(CO)=0.4377m(SO_2)/M(CO)=0.4377n(SO_2)\times M(SO_2)/M(CO)=1.001 \times n(SO_2)}$

即产物中$\mathrm{SO_2}$与$\mathrm{CO}$摩尔数相同，$\mathrm{y=z}$。

<Checkpoint>[根据a. b.重量关系，得出$\mathrm{SO_2}$与$\mathrm{CO}$摩尔数相同](1)</Checkpoint>

实际上根据原子数守恒已经可以解出方程：
$$
\mathrm{x=2y+z}
$$
$$
\mathrm{y=z}
$$
$$
\mathrm{2z=y+8n}
$$

化为最简，方程式为：
$$
24 \mathrm{~N}_{2} \mathrm{O}+8 \mathrm{CS}_{2} \rightarrow 24 \mathrm{~N}_{2}+8 \mathrm{SO}_{2}+8 \mathrm{CO}+\mathrm{S}_{8}
$$

<Checkpoint>[根据原子数守恒解出方程式为$ 24 \mathrm{~N}_{2} \mathrm{O}+8 \mathrm{CS}_{2} \rightarrow 24 \mathrm{~N}_{2}+8 \mathrm{SO}_{2}+8 \mathrm{CO}+\mathrm{S}_{8} $](1)</Checkpoint>

系数乘积为294912，选项B正确。