Salt **A** is a white powder with water of crystallization and plays an important role in the leather industry and dyeing and printing industries. $1.260 \mathrm{~g}$ of salt **A** was heated to $150^{\circ} \mathrm{C}$ in a $\mathrm{N}_{2}$ atmosphere, losing $34.6 \%$ of its weight to obtain **B**, corresponding to the loss of all water of crystallization; **B** was further heated to constant weight, releasing a mixture of two gases in a molar ratio of $1:1$, with a density of $0.61 \mathrm{~g/L}(298.15 \mathrm{~K}, 100 \mathrm{kPa})$; the remaining solid was a white powder **C**, with a mass of $0.642 \mathrm{~g}$. The solutions obtained by dissolving **A**, **B**, and **C** in water all have a certain alkalinity. Which of the following options is correct?
盐 **A** 为白色带结晶水粉末,在皮革工业、印染行业中具有重要的作用。将 $1.260 \mathrm{~g}$ 盐 **A** 置于 $\mathrm{N}_{2}$ 气氛中加热至 $150^{\circ} \mathrm{C}$, 失重 $34.6 \%$ 得到 **B**,对应于失去所有结晶水；对 **B** 继续进行加 热至恒重,放出摩尔比 $1:1$ 的两种气体的混合物,其密度为 $0.61 \mathrm{~g/L}(298.15 \mathrm{~K}, 100 \mathrm{~kPa})$；剩余固体为白色粉末 **C**, 质量为 $0.642 \mathrm{~g}$ 。**A**、**B**、**C** 溶于水所得到的溶液均具有一定的碱性。下列选项正确的是（）
All other options are incorrect
A mixed gas does not necessarily contain $H_{2}$.
Both gases released during thermogravimetric analysis are flammable.
If the reaction is carried out in an air atmosphere, the temperature at which **C** is formed will be higher.
Salt **C** can be used as a leavening agent alone in baking.
Salt **A** contains $1$ crystal water.
其他选项均不正确
混合气体中不一定含有$H_{2}$。
热重释放的两种气体均具有可燃性。
若该反应在空气气氛中进行，则生成 **C** 的温度会更高。
盐 **C** 在烘焙中可单独作为膨松剂。
盐 **A** 中含有$1$个结晶水。
The molar mass of the mixed gas is: $M=\rho R T / p=15.0 \mathrm{~g} / \mathrm{mol}$ ,

<Checkpoint>[Therefore, the mixed gas must contain $\mathrm{H}_{2} $.](1)</Checkpoint>

Since the molar ratio of the two gases is $1: 1$, the molar mass of the other gas is $28 \mathrm{~g} / \mathrm{mol}$,

<Checkpoint>[Which is $\mathrm{CO} $.](1)</Checkpoint>

$n\left(\mathrm{H}_{2}\right)=n(\mathrm{CO})=\left[1.260 \mathrm{~g} \times(1-0.346)-0.642 \mathrm{~g}\right] /[2 \times 15.0 \mathrm{~g} / \mathrm{mol}]=6.07 \times 10^{-3} \mathrm{~mol}$

The number of moles of water of crystallization is $n\left(\mathrm{H}_{2} \mathrm{O}\right)=1.260 \mathrm{~g} \times 0.346 / 18.0 \mathrm{~g} / \mathrm{mol}=2.42 \times 10^{-2} \mathrm{~mol}$

<Checkpoint>[The number of moles of water of crystallization is calculated](1)</Checkpoint>

$n\left(\mathrm{H}_{2} \mathrm{O}\right)=4 n\left(\mathrm{H}_{2}\right)=4 n(\mathrm{CO})$

Suppose there are 2 waters of crystallization, then $M=2 m / n\left(\mathrm{H}_{2} \mathrm{O}\right)=104 \mathrm{~g} / \mathrm{mol}$ minus 2 $\mathrm{H}_{2} \mathrm{O}$, 1/2 $\mathrm{CO}$ and 1/2 $\mathrm{H}_{2}$,

<Checkpoint>[It can be concluded that $\mathrm{A}$ is $\mathrm{HCOONa} \cdot 2 \mathrm{H}_{2} \mathrm{O}$ ](1)</Checkpoint>

The molar mass of the mixed gas is: $M=\rho R T / p=15.0 \mathrm{~g} / \mathrm{mol}$ , The small molecule gases are $H_{2}$, $N_{2}$, $O_{2}$, etc., but only $H_{2}$ has a molar mass within $15.0 \mathrm{~g} / \mathrm{mol}$, so it must contain $H_{2}$.

<Checkpoint>[B is incorrect.](1)</Checkpoint>

The two gases released by thermogravimetry are $H_{2}$ and $CO$, both of which are flammable.

<Checkpoint>[C is correct.](1)</Checkpoint>

If the reaction is carried out in an air atmosphere, the generated gases are all reducing and can react with $\mathrm{O}_{2}$. This reaction is coupled with the decomposition reaction of sodium formate.

<Checkpoint>[The temperature for generating **C** will be lower. D is incorrect.](1)</Checkpoint>

Salt **C** is $Na_2CO_{3}$, which requires the addition of acid to produce $H_{2}O$ and $CO_{2}$. Sodium bicarbonate (commonly known as baking soda) has this unique effect that sodium carbonate (commonly known as soda ash) does not have, that is, it can independently generate gas as a leavening agent without relying on acidic substances to trigger the reaction. It can be used alone as a leavening agent in baking.

<Checkpoint>[Therefore, E is incorrect.](1)</Checkpoint>

Calculated,

<Checkpoint>[Salt **A** contains $2$ waters of crystallization. F is incorrect.](1)</Checkpoint>
混合气体的摩尔质量为:$M=\rho R T / p=15.0 \mathrm{~g} / \mathrm{mol}$  , 

<Checkpoint>[故混合气体中必有 $\mathrm{H}_{2} $。](1)</Checkpoint>

 又因两种气体摩尔比为 $1: 1$, 故另一气体摩尔质量为 $28 \mathrm{~g} / \mathrm{mol}$, 

<Checkpoint>[为 $\mathrm{CO} $。](1)</Checkpoint>

 $n\left(\mathrm{H}_{2}\right)=n(\mathrm{CO})=\left[1.260 \mathrm{~g} \times(1-0.346)-0.642 \mathrm{~g}\right] /[2 \times 15.0 \mathrm{~g} / \mathrm{mol}]=6.07 \times 10^{-3} \mathrm{~mol}$

 结晶水的摩尔数为 $n\left(\mathrm{H}_{2} \mathrm{O}\right)=1.260 \mathrm{~g} \times 0.346 / 18.0 \mathrm{~g} / \mathrm{mol}=2.42 \times 10^{-2} \mathrm{~mol}$ 

<Checkpoint>[计算得到结晶水摩尔数](1)</Checkpoint>

$n\left(\mathrm{H}_{2} \mathrm{O}\right)=4 n\left(\mathrm{H}_{2}\right)=4 n(\mathrm{CO})$

不妨假设有 2 个结晶水, 则 $M=2 m / n\left(\mathrm{H}_{2} \mathrm{O}\right)=104 \mathrm{~g} / \mathrm{mol}$ 减去 2 个 $\mathrm{H}_{2} \mathrm{O}, 1 / 2$ 个 $\mathrm{CO}$ 和 $1 / 2$ 个 $\mathrm{H}_{2}$, 

<Checkpoint>[可得出 $\mathrm{A}$ 为 $\mathrm{HCOONa} \cdot 2 \mathrm{H}_{2} \mathrm{O}$ ](1)</Checkpoint>

由混合气体混合气体的摩尔质量为:$M=\rho R T / p=15.0 \mathrm{~g} / \mathrm{mol}$  , 小分子气体有$H_{2}$、$N_{2}$、$O_{2}$等，但只有$H_{2}$摩尔质量在$15.0 \mathrm{~g} / \mathrm{mol}$以内，所以一定含有$H_{2}$。

<Checkpoint>[B错误。](1)</Checkpoint>

热重释放的两种气体为$H_{2}$和$CO$，均具有可燃性。

<Checkpoint>[C正确。](1)</Checkpoint>

若该反应在空气气氛中进行，生成的气体均具有还原性, 可以与 $\mathrm{O}_{2}$ 反应, 该反应与甲酸钠的分解反应耦合。

<Checkpoint>[生成 **C** 的温度会更低。D错误。](1)</Checkpoint>

盐 **C** 为$Na_2CO_{3}$，需要加酸才能产生$H_{2}O$和$CO_{2}$。碳酸氢钠（俗称小苏打）具有此项独特功效是碳酸钠（俗称纯碱）所不具备的，即作为膨松剂时能独立产生气体，无需依赖酸性物质触发反应。在烘焙中可单独作为膨松剂。

<Checkpoint>[故E错误。](1)</Checkpoint>

计算得到，

<Checkpoint>[盐 **A** 中含有$2$个结晶水。F错误。](1)</Checkpoint>