A new compound $\bf{X}$ was prepared by hydrothermal reaction of $\rm{PbO_2}$ and $\rm{Ag_2O}$ in a silver crucible for 7 days. The following information is related to $\bf{X}$:

(a) 1.000 g of $\bf{X}$ was added to 25.00 mL of 0.1015 mol/L oxalic acid solution acidified, and after sufficient reaction (the existing forms of each element in the product: $\rm Ag \to Ag^+$; $\rm Pb \to Pb^{2+}$) (reaction 1), it was titrated with 0.02030 mol/L acidic $\rm{KMnO_4}$ solution (reaction 2), consuming 21.84 mL.

(b) The lead in 0.500 g of $\bf{X}$ was completely converted to $\rm{PbSO_4}$ precipitate, and its weight was measured to be 0.286 g.

(c) Structure determination showed that there are no hydroxyl groups or water molecules in $\bf{X}$, and the ratio of the number of each atom in $\bf{X}$ is a simple integer ratio; conductivity measurement showed that $\bf{X}$ is a superconductor.

Recently, it has been found that $\bf X$ can also be obtained by mixing and heating $\rm Ag_2O$, $\rm PbO_2$ and a small amount of activated carbon in vacuum (reaction 3).

Let the sum of the stoichiometric numbers of the ionic equations of reactions 1 and 2 be a and b, respectively, and the sum of the stoichiometric numbers of the chemical equation of reaction 3 be c, then (a+b)/c is approximately
在银坩埚中将$\rm{PbO_2}$和$\rm{Ag_2O}$混合水热反应7天，制得了一种新的化合物$\bf{X}$。与$\bf{X}$有关的信息如下：  

(a) 将1.000 g $\bf{X}$ 加入到酸化的 25.00 mL 0.1015 mol/L 的草酸溶液中，充分反应后（产物中各元素存在形式：$\rm Ag \to Ag^+$；$\rm Pb \to Pb^{2+}$）（反应1），用0.02030 mol/L酸性$\rm{KMnO_4}$溶液滴定（反应2），消耗 21.84 mL。  

(b) 将0.500 g $\bf{X}$中的铅完全转化为$\rm{PbSO_4}$沉淀，测得其重量为0.286 g。  

(c) 结构测定表明，$\bf{X}$中不存在羟基或水分子，$\bf{X}$中各原子数目比为简单整数比；导电性测定表明，$\bf{X}$是一种超导体。

最近人们发现，$\bf X$也可由$\rm Ag_2O$、$\rm PbO_2$和少量的活性炭在真空中混合加热反应得到（反应3）。

记反应1、2离子方程式的化学计量数之和分别为a、b，反应3化学方程式的化学计量数之和为c，则(a+b)/c约为
All other options are incorrect
2.67
3.67
1.33
1.67
2.75
3.45
4.09
4.83
3.33
1.88
2.37
3.42
3.50
2.83
2.17
1.92
5.78
6.10
其他选项均不正确
2.67
3.67
1.33
1.67
2.75
3.45
4.09
4.83
3.33
1.88
2.37
3.42
3.50
2.83
2.17
1.92
5.78
6.10
The total amount of electrons transferred by oxalic acid is:

$\rm 2 \times 25.00~mL \times 0.1015~mol/L = 5.075 \times 10^{-3}~mol$

<Checkpoint>[The total amount of electrons transferred by oxalic acid is $\rm 5.075 \times 10^{-3}~mol$](0.5)</Checkpoint>

The total amount of electrons transferred by potassium permanganate is:

$\rm 0.02030~mol/L \times 21.84~mL \times 5 = 2.217 \times 10^{-3}~mol$

<Checkpoint>[The total amount of electrons transferred by potassium permanganate is: $\rm 2.217 \times 10^{-3}~mol$](0.5)</Checkpoint>

So the number of electrons corresponding to 1.000 g of $\bf X$ is:

$\rm 5.075 \times 10^{-3}~mol - 2.217 \times 10^{-3}~mol = 2.858 \times 10^{-3}~mol$

Assume the chemical formula of $\bf X$ is $\rm Ag_xPbO_y$, and assume that $\rm Ag$ and $\rm O$ are +1 and -2 valence, respectively, then:

$\rm (207.2+32.07+16.00 \times 4)/(107.9x+207.2+16.00y)=0.286/0.500~(a)$

$\rm 1.000~g \times (2y-x-2)/(107.9x+207.2+16.00y)=2.858 \times 10^{-3}~mol~(b)$

<Checkpoint>[The two equations are $\rm (207.2+32.07+16.00 \times 4)/(107.9x+207.2+16.00y)=0.286/0.500$ and $\rm 1.000~g \times (2y-x-2)/(107.9x+207.2+16.00y)=2.858 \times 10^{-3}~mol$  ](1)</Checkpoint>

Solving equations (a) and (b) simultaneously yields:

$\rm x=2.54；y = 3.03$

After converting to the simplest integer ratio, we can know that the chemical formula of $\bf X$ is: $\rm Ag_5Pb_2O_6$

<Checkpoint>[The chemical formula of $\bf X$ is: $\rm Ag_5Pb_2O_6$](1)</Checkpoint>

$\rm H_2C_2O_4$ is oxidized to $\rm CO_2$, and the reaction equation is:

Reaction 1: $\rm 2Ag_5Pb_2O_6+18H^++3H_2C_2O_4 \rightarrow 10Ag^++ 4Pb^{2+}+12H_2O+6CO_2$

<Checkpoint>[Reaction 1: $\rm 2Ag_5Pb_2O_6+18H^++3H_2C_2O_4 \rightarrow 10Ag^++ 4Pb^{2+}+12H_2O+6CO_2$](1)</Checkpoint>

Reaction 2: $\rm 2MnO_4^-+5H_2C_2O_4+6H^+ \rightarrow 10CO_2+2Mn^{2+}+8H_2O$

<Checkpoint>[Reaction 2: $\rm 2MnO_4^-+5H_2C_2O_4+6H^+ \rightarrow 10CO_2+2Mn^{2+}+8H_2O$](1)</Checkpoint>

In reaction 3, $\rm Pb$ needs to be reduced, so activated carbon is oxidized to $\rm CO_2$: $\rm 10Ag_2O+8PbO_2+C \rightarrow 4Ag_5Pb_2O_6+CO_2$

<Checkpoint>[Reaction 3: $\rm 10Ag_2O+8PbO_2+C \rightarrow 4Ag_5Pb_2O_6+CO_2$](1)</Checkpoint>

a=55, b=33, c=24, (a+b)/c=3.67, select C
草酸转移电子的总量为：  

$\rm 2 \times 25.00~mL \times 0.1015~mol/L = 5.075 \times 10^{-3}~mol$  

<Checkpoint>[草酸转移电子的总量为$\rm 5.075 \times 10^{-3}~mol$](0.5)</Checkpoint>

高锰酸钾转移电子的总量为：  

$\rm 0.02030~mol/L \times 21.84~mL \times 5 = 2.217 \times 10^{-3}~mol$  

<Checkpoint>[高锰酸钾转移电子的总量为：$\rm 2.217 \times 10^{-3}~mol$](0.5)</Checkpoint>

所以1.000 g $\bf X$对应的电子数为：  

$\rm 5.075 \times 10^{-3}~mol - 2.217 \times 10^{-3}~mol = 2.858 \times 10^{-3}~mol$  

设$\bf X$的化学式为$\rm Ag_xPbO_y$，并假设其中$\rm Ag$和$\rm O$分别为+1价、-2价，则：  

$\rm (207.2+32.07+16.00 \times 4)/(107.9x+207.2+16.00y)=0.286/0.500~(a)$  

$\rm 1.000~g \times (2y-x-2)/(107.9x+207.2+16.00y)=2.858 \times 10^{-3}~mol~(b)$  

<Checkpoint>[列出两个方程组，分别为$\rm (207.2+32.07+16.00 \times 4)/(107.9x+207.2+16.00y)=0.286/0.500$和$\rm 1.000~g \times (2y-x-2)/(107.9x+207.2+16.00y)=2.858 \times 10^{-3}~mol$  ](1)</Checkpoint>

联立(a)、(b)解得：  

$\rm x=2.54；y = 3.03$  

化成最简整数比后可知$\bf X$的化学式为：$\rm Ag_5Pb_2O_6$

<Checkpoint>[$\bf X$的化学式为：$\rm Ag_5Pb_2O_6$](1)</Checkpoint>

$\rm H_2C_2O_4$被氧化为$\rm CO_2$，反应方程式为：  

反应1：$\rm 2Ag_5Pb_2O_6+18H^++3H_2C_2O_4 \rightarrow 10Ag^++ 4Pb^{2+}+12H_2O+6CO_2$  

<Checkpoint>[反应1：$\rm 2Ag_5Pb_2O_6+18H^++3H_2C_2O_4 \rightarrow 10Ag^++ 4Pb^{2+}+12H_2O+6CO_2$](1)</Checkpoint>

反应2：$\rm 2MnO_4^-+5H_2C_2O_4+6H^+ \rightarrow 10CO_2+2Mn^{2+}+8H_2O$

<Checkpoint>[反应2：$\rm 2MnO_4^-+5H_2C_2O_4+6H^+ \rightarrow 10CO_2+2Mn^{2+}+8H_2O$](1)</Checkpoint>

反应3中$\rm Pb$要被还原，故活性炭被氧化为$\rm CO_2$：$\rm 10Ag_2O+8PbO_2+C \rightarrow 4Ag_5Pb_2O_6+CO_2$

<Checkpoint>[反应3：$\rm 10Ag_2O+8PbO_2+C \rightarrow 4Ag_5Pb_2O_6+CO_2$](1)</Checkpoint>

a=55，b=33，c=24，(a+b)/c=3.67，选C