When crystal $\mathrm{A}$ is heated, it generates gas $\mathrm{B}$, solid $\mathrm{C}$, and water. $\mathrm{B}$ reacts with metallic magnesium to produce solid $\mathrm{D}$. $\mathrm{D}$ reacts with water to produce gas $\mathrm{E}$ and precipitate $\mathrm{F}$. An aqueous solution of $\mathrm{A}$ is heated with sodium hydroxide solution to produce gas $\mathrm{E}$ and solution $\mathrm{G}$. The addition of barium chloride solution to $\mathrm{G}$ produces precipitate $\mathrm{H}$. $\mathrm{H}$ is soluble in dilute hydrochloric acid. Given that the thermal decomposition reaction of $\mathrm{A}$ is simplified to the lowest integer ratio, and the coefficient of $\mathrm{A}$ is 1, what is the mass fraction of hydrogen in crystal $\mathrm{A}$?
某晶体$\mathrm{A}$受热时生成气体$\mathrm{B}$，固体$\mathrm{C}$和水。$\mathrm{B}$与金属镁反应生成固体$\mathrm{D}$。$\mathrm{D}$与水作用生成气体$\mathrm{E}$和沉淀$\mathrm{F}$。$\mathrm{A}$的水溶液与氢氧化钠溶液共热生成气体$\mathrm{E}$和溶液$\mathrm{G}$。$\mathrm{G}$中加入氯化钡溶液则生成沉淀$\mathrm{H}$。$\mathrm{H}$可溶于稀盐酸。已知$\mathrm{A}$的热分解反应化到最简整数比，$\mathrm{A}$的系数为1。问氢元素在晶体$\mathrm{A}$中的质量分数是多少？
1.6%
2.6%
2.9%
3.2%
3.4%
5.0%
5.3%
6.3%
6.4%
6.9%
7.5%
8.4%
6.1%
1.6%
2.6%
2.9%
3.2%
3.4%
5.0%
5.3%
6.3%
6.4%
6.9%
7.5%
8.4%
6.1%
According to "Aqueous solution of $\mathrm{A}$ reacts with $NaOH$ upon heating to produce gas $\mathrm{E}$ and solution $\mathrm{G}$", it is known that when ammonium salts (containing $NH_4^+$) are heated with strong bases (such as $NaOH$), ammonia gas ($NH_3$) is produced. Therefore, gas $\mathrm{E}$ is likely to be ammonia gas ($NH_3$).

<Checkpoint>[$\mathrm{E}=NH_3$](0.5)</Checkpoint>

According to "$\mathrm{D}$ reacts with water to produce gas $\mathrm{E}$ and precipitate $\mathrm{F}$", and it has already been deduced that gas $\mathrm{E}$ is ammonia, then $\mathrm{D}$ should be magnesium nitride ($Mg_3N_2$), and $\mathrm{F}$ is magnesium hydroxide ($Mg(OH)_2$).

<Checkpoint>[$\mathrm{D}=Mg_3N_2$](0.5)</Checkpoint>

According to "$\mathrm{B}$ reacts with $Mg$ to produce solid $\mathrm{D}$", and it has already been deduced that $\mathrm{D}$ is magnesium nitride ($Mg_3N_2$), then $\mathrm{B}$ may be nitrogen or ammonia. However, since $\mathrm{E}$ is ammonia, $\mathrm{B}$ is nitrogen ($N_2$).

<Checkpoint>[$\mathrm{B}=N_2$](0.5)</Checkpoint>

According to "$\mathrm{A}$ decomposes upon heating to produce $\mathrm{B}$, $\mathrm{C}$ and water", it can be inferred that $\mathrm{A}$ is an ammonium salt. Since $\mathrm{C}$ is a solid, $\mathrm{A}$ should contain a metal acid radical. Ammonium salts of metal acid radicals that come to mind include ammonium chromate, ammonium dichromate, ammonium vanadate, ammonium molybdate, and ammonium tungstate. The following are discussed separately:

Ammonium chromate or ammonium dichromate: $\mathrm{G}$ is sodium chromate, $\mathrm{H}$ is barium chromate, soluble in dilute hydrochloric acid, which is consistent.

Ammonium vanadate: $\mathrm{G}$ is sodium vanadate, $\mathrm{H}$ is barium vanadate, soluble in dilute hydrochloric acid, which is consistent.

Ammonium molybdate: $\mathrm{G}$ is sodium molybdate, $\mathrm{H}$ is barium molybdate, and the molybdic acid produced by the reaction with dilute hydrochloric acid is slightly soluble, which is inconsistent.

Ammonium tungstate: $\mathrm{G}$ is sodium tungstate, $\mathrm{H}$ is barium tungstate, and the tungstic acid produced by the reaction with dilute hydrochloric acid is insoluble, which is inconsistent.

Possible $\mathrm{A}$ is ammonium chromate, ammonium dichromate, ammonium vanadate, among which only the simplest formula of the thermal decomposition reaction of ammonium dichromate has a coefficient of 1 for $\mathrm{A}$. Therefore, $\mathrm{A}$ is ammonium dichromate ($(NH_4)_2Cr_2O_7$).

<Checkpoint>[$\mathrm{A}=(NH_4)_2Cr_2O_7$](1)</Checkpoint>

The mass fraction of $\mathrm{H}$ is calculated to be 3.199%, choose D.

<Checkpoint>[$w_\mathrm{H}=3.199 ~ \%$](0.5)</Checkpoint>
根据“$\mathrm{A}$的水溶液与$NaOH$共热生成气体$\mathrm{E}$和溶液$\mathrm{G}$”，常见物质中，铵盐（含$NH_4^+$）与强碱（如$NaOH$）共热时，会生成氨气（$NH_3$）。因此，气体$\mathrm{E}$很可能是氨气（$NH_3$）。

<Checkpoint>[$\mathrm{E}=NH_3$](0.5)</Checkpoint>

根据“$\mathrm{D}$与水反应生成气体$\mathrm{E}$和沉淀$\mathrm{F}$”，已经推出气体$\mathrm{E}$为氨气，则$\mathrm{D}$应该为氮化镁（$Mg_3N_2$），$\mathrm{F}$为氢氧化镁（$Mg(OH)_2$）。

<Checkpoint>[$\mathrm{D}=Mg_3N_2$](0.5)</Checkpoint>

根据“$\mathrm{B}$与$Mg$反应生成固体$\mathrm{D}$”，已经推出$\mathrm{D}$为氮化镁（$Mg_3N_2$），则$\mathrm{B}$可能是氮气或者氨气，但$\mathrm{E}$是氨气，所以$\mathrm{B}$是氮气（$N_2$）

<Checkpoint>[$\mathrm{B}=N_2$](0.5)</Checkpoint>

根据“$\mathrm{A}$受热分解生成$\mathrm{B}$、$\mathrm{C}$和水”可以推测$\mathrm{A}$是一种铵盐。因为$\mathrm{C}$是固体，所以$\mathrm{A}$应该含有金属酸根。金属酸根的铵盐可以想到铬酸铵、重铬酸铵、钒酸铵、钼酸铵和钨酸铵。以下分别讨论：

铬酸铵或重铬酸铵：$\mathrm{G}$为铬酸钠，$\mathrm{H}$为铬酸钡，可溶于稀盐酸，符合。

钒酸铵：$\mathrm{G}$为钒酸钠，$\mathrm{H}$为钒酸钡，可溶于稀盐酸，符合。

钼酸铵：$\mathrm{G}$为钼酸钠，$\mathrm{H}$为钼酸钡，与稀盐酸反应产生的钼酸微溶，不符合。

钨酸铵：$\mathrm{G}$为钨酸钠，$\mathrm{H}$为钨酸钡，与稀盐酸反应产生的钨酸难溶，不符合。

可能的$\mathrm{A}$为铬酸铵、重铬酸铵、钒酸铵，其中只有重铬酸铵热分解反应的最简式$\mathrm{A}$的系数为1，因此$\mathrm{A}$为重铬酸铵（$(NH_4)_2Cr_2O_7$）

<Checkpoint>[$\mathrm{A}=(NH_4)_2Cr_2O_7$](1)</Checkpoint>

计算得$\mathrm{H}$的质量分数为3.199%，选D。

<Checkpoint>[$w_\mathrm{H}=3.199 ~ \%$](0.5)</Checkpoint>