A and B are metallic elements, with B being a transition metal, and X is a non-metallic element; A, B, and X form a salt with the chemical formula$A_2BX_4$, which does not contain water of crystallization.$A_2BX_4$ decomposes upon heating in a hydrogen atmosphere, undergoing an endothermic reaction at 365°C; further heating to 515°C results in a weight loss of 12.8%; continued heating to 930°C leads to an additional weight loss of 51.9%, leaving a residual elemental solid. Heating $A_2BX_4$ to 600°C in a hydrogen atmosphere results in a total weight loss of 25.7%, with the remaining solid being the alloy $A_2B$. Calculate the sum of the atomic numbers of the three elements A, B, and X.
A、B 为金属元素，B为过渡金属，X 为非金属元素；A、B、X 形成了化学式为$A_2BX_4$ 的不含结晶水的盐。$A_2BX_4$在氢气气氛下受热分解，在 365°C 时发生一吸热反应；继续加热到 515°C，失重 12.8%；继续加热至 930 °C，再失重 51.9%，残余单质固体。$A_2BX_4$在氢气气氛中加热至 600°C 总失重 25.7%，残余固体为合金 $A_2B$。计算A、B、X三种元素的原子序数之和为多少。
133
160
134
114
142
82
73
105
133
160
134
114
142
82
73
105
From the second equation of weight loss in the problem, the following relationship can be obtained:

(1-25.7%) *[2M(A) + M(B) + 4M(X)] = 2M(A) + M(B)

<Checkpoint>[List the weight loss equation: (1-25.7%) *[2M(A) + M(B) + 4M(X)] = 2M(A) + M(B)](1)</Checkpoint>

From the first equation of weight loss, we get: the remaining elemental metal should be one of A or B. Therefore, list the weight loss equation:

(1-12.8%-51.9%) * [2M(A) + M(B) + 4M(X)] = M(B) or

(1-12.8%-51.9%) * [2M(A) + M(B) + 4M(X)] = 2M(A)

<Checkpoint>[List the weight loss equation: (1-12.8%-51.9%) * [2M(A) + M(B) + 4M(X)] = M(B) or  (1-12.8%-51.9%) * [2M(A) + M(B) + 4M(X)] = 2M(A)](1)</Checkpoint>

Given that X is a non-metallic element, substitute the molecular weights of several common non-metallic elements such as Cl, O, and S to obtain the following results:

When M(X) = 16 :
 M(A) = 48.56, M(B) = 87.91 or
 M(A) = 43.95, M(B) = 97.12

When M(X) = 32 :
 M(A) = 97.12, M(B) = 175.81 or
 M(A) = 87.91, M(B) = 194.2

When M(X) = 35.5 :
 M(A) = 107.74, M(B) = 195.04 or
 M(A) = 97.52, M(B) = 215.49
It is found that when X is Cl and the residual metal is B, there is a reasonable solution.

<Checkpoint>[X=Cl](2)</Checkpoint>

It is calculated from Cl = 35.5 g/mol that the molecular weight of A is 108 and the molecular weight of B is 195. Then A is Ag and B is Pt. Therefore, the sum of the atomic numbers of the three elements is: 47+78+17=142, so choose E.

<Checkpoint>[B=Pt](1)</Checkpoint>
<Checkpoint>[A=Ag](1)</Checkpoint>
<Checkpoint>[47+78+17=142](1)</Checkpoint>
由题目第二个失重方程可得到如下关系式：

(1-25.7%) *[2M(A) + M(B) + 4M(X)] = 2M(A) + M(B)

<Checkpoint>[列出失重方程：(1-25.7%) *[2M(A) + M(B) + 4M(X)] = 2M(A) + M(B)](1)</Checkpoint>

由第一个失重方程得到：剩余单质金属，应为A或B两者之一，由此列出失重方程：

(1-12.8%-51.9%) * [2M(A) + M(B) + 4M(X)] = M(B) 或

(1-12.8%-51.9%) * [2M(A) + M(B) + 4M(X)] = 2M(A)

<Checkpoint>[列出失重方程：(1-12.8%-51.9%) * [2M(A) + M(B) + 4M(X)] = M(B) 或  (1-12.8%-51.9%) * [2M(A) + M(B) + 4M(X)] = 2M(A)](1)</Checkpoint>

已知X为非金属元素，将常见几种非金属元素Cl、O、S等分子量带入，得到下列结果：

当 M(X) = 16  :  
 M(A) = 48.56, M(B) = 87.91或
 M(A) = 43.95, M(B) = 97.12

当 M(X) = 32 :  
 M(A) = 97.12, M(B) = 175.81或
 M(A) = 87.91, M(B) = 194.2

当 M(X) = 35.5 :  
 M(A) = 107.74, M(B) = 195.04或
 M(A) = 97.52, M(B) = 215.49
发现其中X为Cl，残余金属为B时有合理解

<Checkpoint>[X=Cl](2)</Checkpoint>

由Cl=35.5 g/mol 计算出A的分子量为108，B的分子量为195，则A为Ag，B为Pt，由此则得到三种元素的原子序数之和：47+78+17=142，故选择E

<Checkpoint>[B=Pt](1)</Checkpoint>
<Checkpoint>[A=Ag](1)</Checkpoint>
<Checkpoint>[47+78+17=142](1)</Checkpoint>