Hydrated inorganic salt $\bf X$ (chemical formula $\bf MY \rm _ x·yH_2O$) can be obtained by dissolving the element $\bf M$ in an aqueous solution of common inorganic acid $\rm HY$ and crystallizing, where the mass fraction of $\bf M$ is 26.30%. When $\bf X$ is heated, anhydrous $\bf MY \rm _ x·yH_2O$ cannot be obtained. Instead, it loses 50.31% of its weight at $80~^ \circ \rm C$ and transforms into $\bf A$. $\bf A$ loses another 16.77% of its weight at $180~^ \circ \rm C$ (relative to $\bf X$) to obtain black oxide $\bf B$. It is known that the oxidation state of $\bf M$ does not change during the entire weight loss process, and the mixture of gases released during both weight loss events contains water vapor and a colored, irritating gas. Let a be the sum of the stoichiometric coefficients on both sides of the equation for the first weight loss, and b be the sum of the stoichiometric coefficients on both sides of the equation for the second weight loss, then a/b is
水合无机盐$\bf X$(化学式为$\bf MY \rm _ x·yH_2O$)可以由 $\bf M$单质溶于常见无机酸$\rm HY$的水溶液并结晶得到，其中$\bf M$的质量分数26.30%。$\bf X$加热时并不能得到无水的$\bf MY \rm _ x·yH_2O$，而是在$80~^ \circ \rm C$时失重50.31%转为$\bf A$，$\bf A$在$180~^ \circ \rm C$再次失重16.77%(相对于$\bf X$)得到黑色的氧化物$\bf B$。已知在整个失重过程中$\bf M$的氧化态没有发生改变，且两次失重时放出的气体混合物中均含有水蒸气和一种有颜色的刺激性气体。记第一次失重时的方程式左右两侧化学计量数之和为a，第二次失重时的方程式左右两侧化学计量数之和为b，则a/b为（本题中的化学方程式取各物质化学计量数为最简整数比）
All other options are incorrect
1.33
1.47
2.31
4.80
0.94
1.88
2.93
0.87
0.65
0.42
1.28
2.50
3.04
3.33
4.23
3.77
3.53
1.96
其他选项均不正确
1.33
1.47
2.31
4.80
0.94
1.88
2.93
0.87
0.65
0.42
1.28
2.50
3.04
3.33
4.23
3.77
3.53
1.96
From the mass fraction of $\bf M$ and the information on weight loss, it can be seen that the mass fraction of $\bf M$ in oxide $\bf B$ is $\frac{0.2630}{1-0.5031-0.1677}=0.7989$, so the mass fraction of oxygen in $\bf B$ is 20.11%.

If we assume the chemical formula of $\bf B$ is $\bf M \rm O_a$, then the atomic weight of M is $\rm 16a×0.7989/0.2011$

<Checkpoint>[Assume the chemical formula of $\bf B$ is $\bf M \rm O_a$, then the atomic weight of M is $\rm 16a×0.7989/0.2011$](1)</Checkpoint>

Enumeration shows that when $\rm a=1$, $\rm M_{ \bf M} = 63.56~g·mol^{-1}$ , corresponding to $\rm Cu$

<Checkpoint>[$\bf M$ is $\rm Cu$](1)</Checkpoint>

Substituting the mass fraction of $\bf M$ in $\bf X$, we know that the molecular weight of $\bf X$ is 241.63, and the remaining molecular weight after deducting $\bf M$ is about 178.1. Since $\rm H \bf Y$ is a monobasic acid, and considering the common oxidation state of copper, it can be inferred that $\rm x=2$; at this time, since $\bf M$ can be dissolved in $\rm H \bf Y$ to obtain $\bf X$, and a colored irritating gas is released during weight loss, it can be boldly guessed that $\rm H \bf Y$ is nitric acid. Substituting the remaining molecular weight and deducting the formula weight of two nitrate radicals, it can be known that the number of water molecules is exactly 3, which also verifies the guess that $\rm H \bf Y$ is nitric acid. Therefore, $\bf X$ is $\rm Cu(NO_3)_2·3H_2O$

<Checkpoint>[$\bf X$ is $\rm Cu(NO_3)_2·3H_2O$](2)</Checkpoint>

Since the oxidation state of $\rm Cu$ does not change during the weight loss process, the decomposition of $\bf X$ can be regarded as the hydrolysis of copper ions, and heating causes $\bf X$ to lose nitric acid molecules and water molecules, while $\bf A$ is a basic copper nitrate.

From the mass fraction of weight loss, the molecular weight of the first weight loss is $241.63×0.5031=121.56$, and the second is $241.63×0.1677=40.52$ (if you are sensitive enough to the data, you will find that the ratio between the two is very close to 3:1).

The molecular weight of nitric acid is 63.018, and that of water molecule is 18.016. There is no solution in the case of 1 $\rm Cu$, so consider expanding the multiple to 2 $\rm Cu$, weight loss 243.12, corresponding to $\rm 3HNO_3+3H_2O$; and the second is $\rm HNO_3+H_2O$. Therefore, $\bf A$ is $\rm Cu_2(OH) _3NO_3$

<Checkpoint>[$\bf A$ is $\rm Cu_2(OH) _3NO_3$](2)</Checkpoint>

First weight loss: $8\rm [Cu(NO_3)_ 2·3H_2O]=4Cu_2(OH)_3NO_3+3O _2+12NO_2 +18H_2O$

<Checkpoint>[First weight loss: $8\rm [Cu(NO_3)_ 2·3H_2O]=4Cu_2(OH)_3NO_3+3O _2+12NO_2 +18H_2O$](1.5)</Checkpoint>

Second weight loss: $\rm 4Cu_2(OH)_3NO_3=8CuO+O_2+4NO_2+6H_2O$

<Checkpoint>[Second weight loss: $\rm 4Cu_2(OH)_3NO_3=8CuO+O_2+4NO_2+6H_2O$](1.5)</Checkpoint>

$\rm a=45，b=23，a/b=1.96$，select S

<Checkpoint>[$\rm a/b=1.96$](1)</Checkpoint>
由$\bf M$的质量分数和失重信息，可知$\bf M$在氧化物$\bf B$中的质量分数为$\frac{0.2630}{1-0.5031-0.1677}=0.7989$，因此$\bf B$中氧元素的质量分数为20.11%

若设$\bf B$化学式为$\bf M \rm O_a$，则M的原子量为$\rm 16a×0.7989/0.2011$

<Checkpoint>[设$\bf B$化学式为$\bf M \rm O_a$，则M的原子量为$\rm 16a×0.7989/0.2011$](1)</Checkpoint>

枚举可知当$\rm a=1$时， $\rm M_{ \bf M} = 63.56~g·mol^{-1}$ ，对应于$\rm Cu$

<Checkpoint>[$\bf M$为$\rm Cu$](1)</Checkpoint>

代入$\bf X$中$\bf M$的质量分数可知$\bf X$的分子量为241.63，扣除$\bf M$后剩余分子量约为178.1 。由于$\rm H \bf Y$为一元酸，联系铜元素的常见氧化态可以推知$\rm x=2$；此时再由$\bf M$可以溶于$\rm H \bf Y$得到$\bf X$，且失重时放出有颜色的刺激性气体，可以大胆猜测$\rm H \bf Y$为硝酸。代入剩余分子量，扣除两个硝酸根的式量，可知水分子数量恰好为3，同时也验证了$\rm H \bf Y$为硝酸的猜想。因此$\bf X$为$\rm Cu(NO_3)_2·3H_2O$

<Checkpoint>[$\bf X$为$\rm Cu(NO_3)_2·3H_2O$](2)</Checkpoint>

由于失重过程中$\rm Cu$氧化态没有发生变化，因此$\bf X$的分解可以视为铜离子发生水解、加热使$\bf X$脱去了硝酸分子和水分子，而$\bf A$则是一种碱式硝酸铜。

由失重质量分数，第一次失重分子量为$241.63×0.5031=121.56$，第二次为$241.63×0.1677=40.52$（如果对数据足够敏锐的话会发现二者之比非常接近于3:1）。

硝酸的分子量为63.018 ，水分子为 18.016，在1个$\rm Cu$的情况下无解，因此考虑扩倍为2个$\rm Cu$，失重243.12，对应于$\rm 3HNO_3+3H_2O$；而第二次则是$\rm HNO_3+H_2O$。因此，$\bf A$为$\rm Cu_2(OH) _3NO_3$

<Checkpoint>[$\bf A$为$\rm Cu_2(OH) _3NO_3$](2)</Checkpoint>

第一次失重：$8\rm [Cu(NO_3)_ 2·3H_2O]=4Cu_2(OH)_3NO_3+3O _2+12NO_2 +18H_2O$

<Checkpoint>[第一次失重：$8\rm [Cu(NO_3)_ 2·3H_2O]=4Cu_2(OH)_3NO_3+3O _2+12NO_2 +18H_2O$](1.5)</Checkpoint>

第二次失重： $\rm 4Cu_2(OH)_3NO_3=8CuO+O_2+4NO_2+6H_2O$

<Checkpoint>[第二次失重： $\rm 4Cu_2(OH)_3NO_3=8CuO+O_2+4NO_2+6H_2O$](1.5)</Checkpoint>

$\rm a=45，b=23，a/b=1.96$，选S

<Checkpoint>[$\rm a/b=1.96$](1)</Checkpoint>