A pure sample of a certain inorganic salt $\mathbf{X}$ is placed in air and heated to $220 \degree \mathrm{C}$. The thermogravimetric analysis curve shows that the sample begins to lose weight. When $66.7 \%$ (mole fraction) of $\mathbf{X}$ has reacted, the temperature is rapidly increased to $240 \degree \mathrm{C}$, and the remaining substances interact, and the sample loses weight according to another reaction. When heated to constant weight, there is only one inorganic salt in the sample. Continuing to increase the temperature to $315 \degree \mathrm{C}$, the sample enters a new weight loss platform. At the end of the platform, the final sample mass is $86.76 \%$ of the initial mass.

The following information is available:
1. It is known that during the weight loss process, the reduced mass is only removed in the form of $\mathrm{H_2O}$ molecules.
2. After the $315 \degree \mathrm{C}$ weight loss platform ends, the residual solid does not contain the element $\mathrm{H}$.
3. $\mathbf{X}$ contains only one metal element.

Based on the above information, select the correct statement(s):
1. $\mathbf{X}$ is composed of three elements.
2. Four substances are involved in the entire reaction process.
3. After balancing and simplifying the reaction equation for the first heating stage, the coefficients on both sides are the same.
4. The numerical ratio of the metal element to the oxygen element in the chemical formula of the inorganic salt product produced in the second heating stage is $\mathrm{3:10}$.
5. The ratio of the mass of water lost in the second heating stage to the third heating stage is $\mathrm{2:1}$.
将某无机盐 $\mathbf{X}$ 的纯样品置于空气中加热到 $220 \degree \mathrm{C}$ ，从热重分析图上可以观察到样品开始失重。当有 $66.7 \%$ （物质的量分数）的 $\mathbf{X}$ 反应完时，将温度迅速提高至 $240 \degree \mathrm{C}$ ，剩余物质相互作用，样品则会按照另一种反应进行失重。加热至恒重时，样品中仅存在一种无机盐。继续提高温度至 $315 \degree \mathrm{C}$ ，样品开始进入新的失重平台，平台结束时，最终样品质量为初始质量的 $86.76 \%$ 。

现有以下信息：
1. 已知失重过程中，减少的质量仅以 $\mathrm{H_2O}$ 分子形式脱去。  
2. $315 \degree \mathrm{C}$ 失重平台结束后，残余固体中不含有 $\mathrm{H}$ 元素。  
3. $\mathbf{X}$ 中仅含有一种金属元素。

请根据上述信息，选出正确的说法：
1. $\mathbf{X}$ 由三种元素组成。
2. 在整个反应过程中，涉及到四种物质。
3. 第一个加热阶段的反应方程式经配平化简后，两边的系数相同。
4. 第二个加热阶段生成的无机盐产物化学式中金属元素与氧元素的数值比为 $\mathrm{3:10}$ 。
5. 第二个加热阶段与第三个加热阶段下失水质量比为 $\mathrm{2:1}$ 。
1
1, 4
1, 5
2, 3
2, 4
3, 4
2, 3, 5
3, 4, 5
4, 5
4
5
All of the above options are incorrect.
1
1, 4
1, 5
2, 3
2, 4
3, 4
2, 3, 5
3, 4, 5
4, 5
4
5
以上选项均不正确
Let the chemical formula molar mass of $\mathbf{X}$ be $\mathrm{M(X)~g/mol}$, and the initial amount of substance be $\mathrm{n~mol}$. The final product is represented as $\mathbf{Y}$.

Initial mass: $\mathrm{m(X) = n \times M(X)}$.

Final mass: $\mathrm{m(Z) = 0.8676 \times m(X) = 0.8676 \times n \times  M(X)}$.

Total mass loss ($\mathrm{H_2O}$): $\mathrm{\Delta m = (1 - 0.8676) \times n  \times M(X) = 0.1324 \times n \times M(X)}$

According to the law of conservation of mass, the mass loss is equal to the mass of water removed. Let each molecule of $\mathbf{X}$ ultimately lose $\mathrm{k}$ molecules of $\mathrm{H_2O}$.

$\mathrm{\Delta m = n \times k \times M(H_2O) = n \times k \times 18}$

Establish the equation: $\mathrm{n \times k \times 18 = 0.1324 \times n \times M(X) }$

Obtain:
$\mathrm{M(X) = (18 \times k) / 0.1324 ≈ 135.95 \times k}$

<Checkpoint>[$\mathrm{M(X) = (18 \times k) / 0.1324 ≈ 135.95 \times k}$](1)</Checkpoint>

If $\mathrm{k=1}$, $\mathrm{M(X) ≈ 135.95 \times 1 = 136.0~g/mol}$;

If $\mathrm{k=2}$, $\mathrm{M(X) ≈ 135.95 \times 2 = 271.9~g/mol}$;

If $\mathrm{k=3}$, $\mathrm{M(X) ≈ 135.95 \times 3 = 407.9~g/mol}$

First try the case of $\mathrm{k=1}$. At this time, $\mathrm{M(X) ≈ 136.0~g/mol}$, each molecule of $\mathbf{X}$ ultimately loses $\mathrm{1}$ molecule of $\mathrm{H_2O}$. It is known that the final product $\mathbf{Y}$ does not contain the element $\mathrm{H}$, it can be known that the molecular formula of $\mathbf{X}$ contains $\mathrm{2}$ $\mathrm{H}$ atoms, and the chemical formula molar mass of $\mathbf{Y}$ is $\mathrm{M(Y) = M(X) -18 =118.0~g/mol}$.

<Checkpoint>[When $\mathrm{k=1}$, the chemical formula molar mass of $\mathbf{X}$ is $\mathrm{M(X) =136.0~g/mol}$](1)</Checkpoint>

<Checkpoint>[When $\mathrm{k=1}$, the chemical formula molar mass of $\mathbf{Y}$ is $\mathrm{M(Y) =118.0~g/mol}$](1)</Checkpoint>

It is known that $\mathbf{X}$ contains only one metal element and has $\mathrm{2}$ $\mathrm{H}$ atoms, the molar mass of the remaining acid radical is $\mathrm{136.0-2-M(M)~g/mol}$;
Common metals include alkali metals, alkaline earth metals, etc.; According to the above information, we can exhaustively try:
When $\mathrm{M = K}$, the acid radical is $\mathrm{PO_4^{3-}}$; The molar mass is $\mathrm{93~g/mol}$. $\mathbf{X}$ is $\mathrm{KH_2PO_4}$; and $\mathbf{Y}$ is $\mathrm{KPO_3}$; $\mathrm{M(Y) = 118.0~g/mol}$.

The inorganic salt in the question undergoes $\mathrm{3}$ mass loss platforms, indicating that $\mathrm{H_2O}$ is lost in all three processes; Let the inorganic salt generated on the first platform be $\mathbf{P1}$; the inorganic salt generated on the second platform be $\mathbf{P2}$; the final product generated on the third platform is $\mathbf{Y}$.

According to the question, the first platform consumes $\mathrm{\frac{2}{3}n}$ $\mathbf{X}$ to generate $\mathbf{P1}$, and on the second platform $\mathrm{\frac{1}{3}n}$ $\mathbf{X}$ just reacts completely with $\mathbf{P1}$ to generate $\mathbf{P2}$. It can be known that $\mathrm{\frac{1}{3}n}$ $\mathbf{P1}$ is generated on the first platform, and $\mathrm{\frac{1}{3}n}$ $\mathbf{P2}$ is generated on the second platform. That is, the coefficient ratio of the first reaction equation: $\mathbf{X}$ : $\mathbf{P1}$ = $\mathrm{2:1}$; the coefficient ratio of the second reaction equation: $\mathbf{X}$ : $\mathbf{P1}$ : $\mathbf{P2}$ = $\mathrm{1:1:1}$.

<Checkpoint>[The coefficient ratio of the first reaction equation: $\mathbf{X}$ : $\mathbf{P1}$ = $\mathrm{2:1}$](1)</Checkpoint>

<Checkpoint>[The coefficient ratio of the second reaction equation: $\mathbf{X}$ : $\mathbf{P1}$ : $\mathbf{P2}$ = $\mathrm{1:1:1}$](1)</Checkpoint>

According to the above information, try to write the mass loss equation of $\mathrm{KH_2PO_4}$:

$\mathrm{2KH_2PO_4=H_2O+K_2H_2P_2O_7}$;

$\mathrm{KH_2PO_4+K_2H_2P_2O_7=H_2O+K_3H_2P_3O_{10}}$;

$\mathrm{K_3H_2P_3O_{10}=H_2O+3KPO_3}$;

The overall equation is:

$\mathrm{KH_2PO_4=H_2O+KPO_3}$.

<Checkpoint>[The first reaction equation: $\mathrm{2KH_2PO_4=H_2O+K_2H_2P_2O_7}$](2)</Checkpoint>

<Checkpoint>[The second reaction equation: $\mathrm{KH_2PO_4+K_2H_2P_2O_7=H_2O+K_3H_2P_3O_{10}}$](2)</Checkpoint>

<Checkpoint>[The third reaction equation: $\mathrm{K_3H_2P_3O_{10}=H_2O+3KPO_3}$](2)</Checkpoint>

<Checkpoint>[The overall reaction equation: $\mathrm{KH_2PO_4=H_2O+KPO_3}$](2)</Checkpoint>

This process conforms to the question. Therefore, $\mathbf{X}$ is $\mathrm{KH_2PO_4}$; $\mathbf{Y}$ is $\mathrm{KPO_3}$.

<Checkpoint>[$\mathbf{X}$ is $\mathrm{KH_2PO_4}$](2)</Checkpoint>

<Checkpoint>[$\mathbf{Y}$ is $\mathrm{KPO_3}$](2)</Checkpoint>

Therefore, statements 3 and 4 are correct, choose option F.
设 $\mathbf{X}$ 的化学式摩尔质量为 $\mathrm{M(X)~g/mol}$ ，初始物质的量为 $\mathrm{n~mol}$ 。最终产物表示为 $\mathbf{Y}$ 。

初始质量： $\mathrm{m(X) = n \times M(X)}$ 。

最终质量：$\mathrm{m(Z) = 0.8676 \times m(X) = 0.8676 \times n \times  M(X)}$ 。

失重总质量（$\mathrm{H_2O}$）：$\mathrm{\Delta m = (1 - 0.8676) \times n  \times M(X) = 0.1324 \times n \times M(X)}$

根据质量守恒，失重质量等于脱去水的质量。设每分子 $\mathbf{X}$ 最终脱去 $\mathrm{k}$ 个 $\mathrm{H_2O}$ 分子。 

$\mathrm{\Delta m = n \times k \times M(H_2O) = n \times k \times 18}$

建立等式：$\mathrm{n \times k \times 18 = 0.1324 \times n \times M(X) }$

求得：
$\mathrm{M(X) = (18 \times k) / 0.1324 ≈ 135.95 \times k}$

<Checkpoint>[$\mathrm{M(X) = (18 \times k) / 0.1324 ≈ 135.95 \times k}$](1)</Checkpoint>

若 $\mathrm{k=1}$ ，$\mathrm{M(X) ≈ 135.95 \times 1 = 136.0~g/mol}$；

若 $\mathrm{k=2}$ ，$\mathrm{M(X) ≈ 135.95 \times 2 = 271.9~g/mol}$；

若 $\mathrm{k=3}$，$\mathrm{M(X) ≈ 135.95 \times 3 = 407.9~g/mol}$

先尝试 $\mathrm{k=1}$ 的情况。此时，$\mathrm{M(X) ≈ 136.0~g/mol}$，每分子 $\mathbf{X}$ 最终脱去 $\mathrm{1}$ 个 $\mathrm{H_2O}$ 分子。 已知最终产物 $\mathbf{Y}$ 不含 $\mathrm{H}$ 元素，可知 $\mathbf{X}$ 的分子式含有 $\mathrm{2}$ 个 $\mathrm{H}$ 原子，且 $\mathbf{Y}$ 的化学式摩尔质量为 $\mathrm{M(Y) = M(X) -18 =118.0~g/mol}$。

<Checkpoint>[当 $\mathrm{k=1}$时，$\mathbf{X}$ 的化学式摩尔质量为 $\mathrm{M(X) =136.0~g/mol}$](1)</Checkpoint>

<Checkpoint>[当 $\mathrm{k=1}$时，$\mathbf{Y}$ 的化学式摩尔质量为 $\mathrm{M(Y) =118.0~g/mol}$](1)</Checkpoint>

已知 $\mathbf{X}$ 中只含一种金属元素且有 $\mathrm{2}$ 个 $\mathrm{H}$ 原子，剩下酸根的摩尔质量为 $\mathrm{136.0-2-M(M)~g/mol}$；
常见的金属有碱金属、碱土金属等；根据以上信息，可以穷举试出：
当 $\mathrm{M = K}$ 时，酸根为 $\mathrm{PO_4^{3-}}$ ；摩尔质量为 $\mathrm{93~g/mol}$ 。$\mathbf{X}$ 为 $\mathrm{KH_2PO_4}$ ；且 $\mathbf{Y}$ 为 $\mathrm{KPO_3}$ ；$\mathrm{M(Y) = 118.0~g/mol}$ 。

题中该无机盐经过 $\mathrm{3}$ 个失重平台，表示三个过程都有  $\mathrm{H_2O}$ 丢失；设第一个平台生成无机盐 $\mathbf{P1}$ ；第二个平台生成无机盐 $\mathbf{P2}$ ；第三个平台生成最终产物 $\mathbf{Y}$ 。

根据题意，第一个平台消耗 $\mathrm{\frac{2}{3}n}$ $\mathbf{X}$ 生成 $\mathbf{P1}$ ，第二个平台 $\mathrm{\frac{1}{3}n}$ $\mathbf{X}$ 刚好与 $\mathbf{P1}$ 反应完全生成  $\mathbf{P2}$ 。可知在第一个平台生成了 $\mathrm{\frac{1}{3}n}$ $\mathbf{P1}$，第二个平台生成了 $\mathrm{\frac{1}{3}n}$ $\mathbf{P2}$。即第一个反应方程式系数比： $\mathbf{X}$ ：$\mathbf{P1}$ = $\mathrm{2:1}$；第二个反应方程式系数比： $\mathbf{X}$ ：$\mathbf{P1}$ ：$\mathbf{P2}$ = $\mathrm{1:1:1}$。

<Checkpoint>[第一个反应方程式系数比： $\mathbf{X}$ ：$\mathbf{P1}$ = $\mathrm{2:1}$](1)</Checkpoint>

<Checkpoint>[第二个反应方程式系数比： $\mathbf{X}$ ：$\mathbf{P1}$ ：$\mathbf{P2}$ = $\mathrm{1:1:1}$](1)</Checkpoint>

根据以上信息，尝试书写 $\mathrm{KH_2PO_4}$失重方程式：

$\mathrm{2KH_2PO_4=H_2O+K_2H_2P_2O_7}$；

$\mathrm{KH_2PO_4+K_2H_2P_2O_7=H_2O+K_3H_2P_3O_{10}}$；

$\mathrm{K_3H_2P_3O_{10}=H_2O+3KPO_3}$；

总方程式为：

$\mathrm{KH_2PO_4=H_2O+KPO_3}$。

<Checkpoint>[第一个反应方程式：$\mathrm{2KH_2PO_4=H_2O+K_2H_2P_2O_7}$](2)</Checkpoint>

<Checkpoint>[第二个反应方程式：$\mathrm{KH_2PO_4+K_2H_2P_2O_7=H_2O+K_3H_2P_3O_{10}}$](2)</Checkpoint>

<Checkpoint>[第三个反应方程式：$\mathrm{K_3H_2P_3O_{10}=H_2O+3KPO_3}$](2)</Checkpoint>

<Checkpoint>[总反应方程式：$\mathrm{KH_2PO_4=H_2O+KPO_3}$](2)</Checkpoint>

该过程符合题设。因此 $\mathbf{X}$ 为 $\mathrm{KH_2PO_4}$ ；$\mathbf{Y}$ 为 $\mathrm{KPO_3}$ 。

<Checkpoint>[$\mathbf{X}$ 为 $\mathrm{KH_2PO_4}$](2)</Checkpoint>

<Checkpoint>[$\mathbf{Y}$ 为 $\mathrm{KPO_3}$](2)</Checkpoint>

因此说法3, 4正确，选择选项 F 。