Fluorine gas reacts with hydrazoic acid to produce 3 products: a certain pseudohalogen-halogen interhalide, a gas, and a common salt. Which of the following statements is correct:
氟气与叠氮酸发生反应，生成 3 种产物：某种拟卤素-卤素互化物、气体与一种常见盐。下列说法正确的是：
The generated pseudohalogen-halogen interhalide is $\mathrm{N}_{3} \mathrm{F}_{3}$
Gaseous product is ammonia.
The stoichiometric ratio of fluorine gas to hydrazoic acid in this reaction may be 1:4
The gaseous product is nitrogen trifluoride.
For every mole of fluorine gas consumed in the reaction, 6 moles of gas may be produced.
The anion in the salt produced by the reaction is $\mathrm{N}_3^-$
All other options are incorrect.
生成的拟卤素-卤素互化物为 $\mathrm{N}_{3} \mathrm{F}_{3}$
气体产物为氨
该反应中氟气与叠氮酸的化学计量数之比可能为 1:4
气体产物为三氟化氮
反应每消耗1摩尔氟气可能生成 6 摩尔气体
反应生成的盐中阴离子为 $\mathrm{N}_3^-$
其他选项均不正确
The only halogen in the question is fluorine, and the only pseudohalogen is azide, so the pseudohalogen-halogen interhalogen compound is fluorine azide.

<Checkpoint>[The pseudohalogen-halogen interhalogen compound is $\mathrm { F N } _ { 3 }$](1)</Checkpoint>

Azide is oxidized to nitrogen gas by strongly oxidizing fluorine, so the gas is nitrogen gas.

<Checkpoint>[The gaseous product is ${\mathrm{N}_{2}}$](1)</Checkpoint>

The hydrogen element can only exist as ammonium ion, and the anion is the reduction product fluoride ion, so the salt is ammonium fluoride.

<Checkpoint>[The salt product is ${\mathrm{NH}_{4}\mathrm{F}}$](1)</Checkpoint>

Therefore, options A, B, and D are all incorrect. Next, balance the equation. There are two reasonable equations:

$$ { \mathrm { F } _ { 2 } + 4 \mathrm { H N } _ { 3 } \to\mathrm { F N } _ { 3 } + \mathrm { N H } _ { 4 } }\mathrm{F} +\mathrm{4N}_{2}$$

$$ { \mathrm {2 F } _ { 2 } + 4 \mathrm { H N } _ { 3 } \to\mathrm { 3F N } _ { 3 } + \mathrm { N H } _ { 4 } }\mathrm{F} +\mathrm{N}_{2}$$

<Checkpoint>[$$ { \mathrm { F } _ { 2 } + 4 \mathrm { H N } _ { 3 } \to\mathrm { F N } _ { 3 } + \mathrm { N H } _ { 4 } }\mathrm{F} +\mathrm{4N}_{2}$$ $$ { \mathrm {2 F } _ { 2 } + 4 \mathrm { H N } _ { 3 } \to\mathrm { 3F N } _ { 3 } + \mathrm { N H } _ { 4 } }\mathrm{F} +\mathrm{N}_{2}$$](2)</Checkpoint>

Fluorine azide is also a gas, so the stoichiometric ratio of fluorine gas to gas products is between 1:2 and 1:5.

<Checkpoint>[The stoichiometric ratio of fluorine gas to the produced gases is 1:2 to 1:5](1)</Checkpoint>

E is incorrect.

The stoichiometric ratio of fluorine gas to hydrazoic acid is between 1:2 and 1:4.

<Checkpoint>[The stoichiometric ratio of fluorine gas to hydrazoic acid is 1:2 to 1:4](1)</Checkpoint>

C is correct.
题中的卤素只有氟，而拟卤素只有叠氮，因此拟卤素-卤素互化物为叠氮化氟。

<Checkpoint>[拟卤素-卤素互化物为 $\mathrm { F N } _ { 3 }$](1)</Checkpoint>

叠氮会被强氧化性的氟氧化为氮气，因此气体为氮气。

<Checkpoint>[气体产物为 ${\mathrm{N}_{2}}$](1)</Checkpoint>

氢元素只能以铵根离子存在，而阴离子为还原产物氟离子，因此盐为氟化铵。

<Checkpoint>[盐产物为 ${\mathrm{NH}_{4}\mathrm{F}}$](1)</Checkpoint>

因此选项A,B,D均错误。接下来配平方程式，合理的方程式有两个：

$$ { \mathrm { F } _ { 2 } + 4 \mathrm { H N } _ { 3 } \to\mathrm { F N } _ { 3 } + \mathrm { N H } _ { 4 } }\mathrm{F} +\mathrm{4N}_{2}$$

$$ { \mathrm {2 F } _ { 2 } + 4 \mathrm { H N } _ { 3 } \to\mathrm { 3F N } _ { 3 } + \mathrm { N H } _ { 4 } }\mathrm{F} +\mathrm{N}_{2}$$

<Checkpoint>[$$ { \mathrm { F } _ { 2 } + 4 \mathrm { H N } _ { 3 } \to\mathrm { F N } _ { 3 } + \mathrm { N H } _ { 4 } }\mathrm{F} +\mathrm{4N}_{2}$$ $$ { \mathrm {2 F } _ { 2 } + 4 \mathrm { H N } _ { 3 } \to\mathrm { 3F N } _ { 3 } + \mathrm { N H } _ { 4 } }\mathrm{F} +\mathrm{N}_{2}$$](2)</Checkpoint>

叠氮化氟也是气体，因此氟气与生成物中气体的化学计量比在1:2与1:5之间。

<Checkpoint>[氟气与生成气体的计量比1:2至1:5](1)</Checkpoint>

E错误。

氟气与叠氮酸的化学计量比在1:2与1:4之间。

<Checkpoint>[氟气与叠氮酸的化学计量比1:2至1:4](1)</Checkpoint>

C正确。