$\rm V_2O_5$ and $\rm S_2Cl_2$ react at a 1:3 ratio at high temperature to produce three substances.

Select the correct option regarding the balanced chemical reaction equation.
$\rm V_2O_5$和$\rm S_2Cl_2$在高温下1:3反应生成三种物质。

选择关于配平的化学反应方程式的正确选项。
All other options are incorrect
The product of all coefficients in a chemical reaction equation is 108.
The product of all coefficients in a chemical reaction equation is 1860.
The molar ratio of sulfur-containing products in the reaction products is 1:1
Vanadium-containing reaction product partial hydrolysis will generate vanadium-oxygen double bonds.
The coefficient of vanadium pentoxide in the reaction equation is 1.
其他选项均不正确
化学反应方程式中所有系数的乘积是108
化学反应方程式中所有系数的乘积是1860
反应产物中含硫产物的摩尔比是1:1
含钒反应产物部分水解将生成钒-氧双键
反应方程式中五氧化二钒的系数是1
Vanadium pentoxide has strong oxidizing properties and can oxidize +1 valent sulfur to sulfur dioxide at high temperatures. With a reactant ratio of 1:3, vanadium is insufficient to oxidize all sulfur to sulfur dioxide, inevitably leaving some sulfur in a lower valence state, the most reasonable form of which is elemental sulfur. Disulfur dichloride can act as a chlorinating agent, so the products should be vanadium chloride, sulfur dioxide, and elemental sulfur.

Among vanadium chlorides, $VCl_3$ and $VCl_4$ are relatively stable. Since there is an excess of low-valent sulfur present here, the lower valence state $VCl_3$ is more reasonable.

<Checkpoint>[The products are $SO_2$, $VCl_3$, and $S$.](1)</Checkpoint>

The balanced chemical reaction equation is:

$$
2 V_2O_5 + 6 S_2Cl_2 \rightarrow 4VCl_3 + 5SO_2 + 7S
$$

<Checkpoint>[$$ 2 V_2O_5 + 6 S_2Cl_2 \rightarrow 4VCl_3 + 5SO_2 + 7S $$](2)</Checkpoint>

The product of all the coefficients in the chemical reaction equation is 1680. Therefore, options B and C are incorrect.

The molar ratio of sulfur-containing products is 5:7. Therefore, option D is incorrect.

Partial hydrolysis of vanadium trichloride cannot generate vanadium-oxygen double bonds. Therefore, option E is incorrect.

<Checkpoint>[Partial hydrolysis of $VCl_3$ cannot generate vanadium-oxygen double bonds](1)</Checkpoint>

The coefficient of vanadium pentoxide is 2. Therefore, option F is incorrect.

There is no correct option, so choose option A.
五氧化二钒有较强氧化性，在高温下可将+1价的硫氧化为二氧化硫。在1:3的反应物比例下，钒不足以将所有硫氧化至二氧化硫，必然有一部分硫为低价态，最合理存在形式即为硫单质。二氯化二硫可作为氯化剂，因此产物应当生成钒的氯化物、二氧化硫、硫单质。

钒的氯化物中$VCl_3$、$VCl_4$较稳定。而此处有过量的低价硫存在，因此价态较低的$VCl_3$更合理。

<Checkpoint>[产物为$SO_2$、$VCl_3$、$S$。](1)</Checkpoint>

配平的化学反应方程式为：

$$
2 V_2O_5 + 6 S_2Cl_2 \rightarrow 4VCl_3 + 5SO_2 + 7S
$$

<Checkpoint>[$$ 2 V_2O_5 + 6 S_2Cl_2 \rightarrow 4VCl_3 + 5SO_2 + 7S $$](2)</Checkpoint>

化学反应方程式中所有系数的乘积是1680。故选项B、C错误。

含硫产物的摩尔比为5:7。故选项D错误。

三氯化钒部分水解不能生成钒-氧双键。故选项E错误。

<Checkpoint>[$VCl_3$部分水解不能生成钒-氧双键](1)</Checkpoint>

五氧化二钒的系数是2。故选项F错误。

没有正确选项，故选择选项A。