Chromium is a steel-gray metal extracted from red Siberian ore by the French scientist Vauquelin in 1797. It is often used in the manufacture of stainless steel and for chrome plating metal surfaces of instruments. Industrially, the preparation of elemental chromium starts from chromite ($\mathrm{FeCr_2O_4}$), which is strongly heated in air with a mixture of sodium carbonate, where iron is converted to ferric oxide (Reaction 1). The resulting solid residue is leached with water, acidified, and a chromium-containing sodium salt precipitates. This salt is then reduced to chromium oxide by heating with carbon at high temperature, simultaneously generating a salt and releasing a gas (Reaction 2). Finally, treatment with ignited metallic aluminum yields relatively pure chromium (Reaction 3).

For the extraction of molybdenum, which is in the same group as chromium, molybdenite is usually used as the raw material, which is roasted in oxygen (Reaction 4). The resulting product is leached with ammonia water to transfer molybdenum to the solution; alternatively, molybdenite can react with caustic soda solution in an oxygen atmosphere to transfer molybdenum to the solution (Reaction 5), during which sulfur is also oxidized and enters the solution. Subsequently, the ammonia-leached solution is acidified to precipitate a polyoxometalate which anion contains seven molybdenum atoms (Reaction 6, ionic equation), which is then calcined and reduced with hydrogen to obtain relatively pure metallic molybdenum.

Let the sum of the coefficients of reactants and products in the equations of Reactions 1 to 6 (requiring the coefficient ratio to be the simplest integer ratio) be $a_i,i=1,2,3,4,5,6$, and the absolute value of the difference between the coefficients of reactants and products be $b_i,i=1,2,3,4,5,6$, then the value of $\log_{\frac{E(a)}{D(a)}}\frac{E(b)}{D(b)}$ is (retain two decimal places in the calculation process, and the final result error within 0.01 is considered correct; Here, $\rm D$ is calculated using the population standard deviation):
铬是由法国人Vauquelin于1797年从红色西伯利亚矿石中提取出的钢灰色金属，它可常用于制造不锈钢和仪器的金属表面镀铬。工业上制备铬单质从铬铁矿($\mathrm{FeCr_2O_4}$)出发，将其与碳酸钠的混合物在空气中强加热，其中铁转化成三氧化二铁(反应1)。将反应后的固体残渣用水浸取，酸化，析出含铬的钠盐。使该盐与碳在高温下共热还原为铬氧化物，同时生成一种盐并放出气体(反应2)。最后，用金属铝点燃处理便可得较纯的铬(反应3)。

对于与铬同族的钼的提取，通常是以辉钼矿作为原料，在氧气下焙烧(反应4)，所得产物用氨水浸出，使钼转移至溶液中；或者，使辉钼矿在氧气气氛中与苛性钠溶液反应，也可使钼转移到溶液中(反应5)，该过程中硫也氧化而进入溶液。随后将氨水浸出的溶液酸化，析出一种阴离子含有七个钼的同多酸盐(反应6，离子方程式)，之后经焙解、氢气还原，便可得较纯的金属钼。

记反应1~反应6的方程式(要求系数比为最简整数比)反应物与生成物系数之和分别为$a_i,i=1,2,3,4,5,6$，反应物与生成物系数之差的绝对值分别为$b_i,i=1,2,3,4,5,6$，则$\log_{\frac{E(a)}{D(a)}}\frac{E(b)}{D(b)}$的值为（计算过程中保留两位小数，最终结果误差在0.01之内视为正确；这里的$\rm D$用总体标准差进行计算）：
All other options are incorrect
-0.98
-0.83
-0.69
-0.52
-0.35
-0.12
0.17
0.33
0.64
0.81
1.02
其他选项均不正确
-0.98
-0.83
-0.69
-0.52
-0.35
-0.12
0.17
0.33
0.64
0.81
1.02
The solid obtained after reaction 1 can be leached with water to obtain a chromium-containing sodium salt. Therefore, reaction 1 should produce +6 valent chromium. Thus, reaction 1 is:

$\mathrm{4FeCr_2O_4+8Na_2CO_3+7O_2\to 2Fe_2O_3+8Na_2CrO_4+8CO_2}$

<Checkpoint>[Reaction 1:$\mathrm{4FeCr_2O_4+8Na_2CO_3+7O_2\to 2Fe_2O_3+8Na_2CrO_4+8CO_2}$](1)</Checkpoint>

After leaching the product with water and acidifying it, sodium chromate is converted to sodium dichromate. Sodium dichromate is reduced to chromium oxide, which should be chromium(III) oxide, when co-heated with carbon at high temperature, and the gas should be carbon monoxide. Therefore, reaction 2 is:

$\mathrm{Na_2Cr_2O_7+2C\to Cr_2O_3+Na_2CO_3+CO}$

<Checkpoint>[Reaction 2:$\mathrm{Na_2Cr_2O_7+2C\to Cr_2O_3+Na_2CO_3+CO}$](1)</Checkpoint>

The final step is a displacement reaction involving aluminum, and the equation is:

$\mathrm{Cr_2O_3+2Al\to Al_2O_3+2Cr}$

<Checkpoint>[Reaction 3:$\mathrm{Cr_2O_3+2Al\to Al_2O_3+2Cr}$](1)</Checkpoint>

The main component of molybdenite is $\mathrm{MoS_2}$, and the equation for its roasting in oxygen is:

$\mathrm{2MoS_2+7O_2\to 2MoO_3+4SO_2}$

<Checkpoint>[Reaction 4:$\mathrm{2MoS_2+7O_2\to 2MoO_3+4SO_2}$](1)</Checkpoint>

When molybdenite reacts with sodium hydroxide solution in an oxygen atmosphere, the sulfur element can be oxidized to +6 valence. Therefore, the equation is:

$\mathrm{2MoS_2+9O_2+12NaOH\to 2Na_2MoO_4+4Na_2SO_4+6H_2O}$

<Checkpoint>[Reaction 5:$\mathrm{2MoS_2+9O_2+12NaOH\to 2Na_2MoO_4+4Na_2SO_4+6H_2O}$](1)</Checkpoint>

Reaction 6 produces a polymolybdate containing 7 molybdenum atoms, and its equation is:

$\mathrm{7MoO_4^{2-}+8H^++6NH_4^+\to (NH_4)_6Mo_7O_{24}+4H_2O}$

<Checkpoint>[Reaction 6:$\mathrm{7MoO_4^{2-}+8H^++6NH_4^+\to (NH_4)_6Mo_7O_{24}+4H_2O}$](1)</Checkpoint>

Therefore:

||1|2|3|4|5|6|
|---|---|---|---|---|---|---|
|$a$|37|6|6|15|35|26|
|$b$|1|0|0|3|11|16|

$E(a)=20.83,D(a)=12.67,E(b)=5.17,D(b)=6.15$

$\frac{E(a)}{D(a)}=1.64,\frac{E(b)}{D(b)}=0.84$

<Checkpoint>[$\frac{E(a)}{D(a)}=1.64$](0.5)</Checkpoint>

<Checkpoint>[$\frac{E(b)}{D(b)}=0.84$](0.5)</Checkpoint>

$\log_{\frac{E(a)}{D(a)}}\frac{E(b)}{D(b)}=-0.35$
反应1反应后得到的固体用水浸取可以得到含铬的钠盐，因此反应1应该产生了+6价的铬，因此反应1为：

$\mathrm{4FeCr_2O_4+8Na_2CO_3+7O_2\to 2Fe_2O_3+8Na_2CrO_4+8CO_2}$

<Checkpoint>[反应1:$\mathrm{4FeCr_2O_4+8Na_2CO_3+7O_2\to 2Fe_2O_3+8Na_2CrO_4+8CO_2}$](1)</Checkpoint>

产物用水浸取酸化后铬酸钠转化为重铬酸钠，重铬酸钠与碳在高温下共热被还原为铬氧化物应为三氧化二铬，气体应为一氧化碳，因此反应2为：

$\mathrm{Na_2Cr_2O_7+2C\to Cr_2O_3+Na_2CO_3+CO}$

<Checkpoint>[反应2:$\mathrm{Na_2Cr_2O_7+2C\to Cr_2O_3+Na_2CO_3+CO}$](1)</Checkpoint>

最后一步为铝发生的置换反应，方程式为：

$\mathrm{Cr_2O_3+2Al\to Al_2O_3+2Cr}$

<Checkpoint>[反应3:$\mathrm{Cr_2O_3+2Al\to Al_2O_3+2Cr}$](1)</Checkpoint>

辉钼矿主要成分为$\mathrm{MoS_2}$，其在氧气下焙烧的方程式为：

$\mathrm{2MoS_2+7O_2\to 2MoO_3+4SO_2}$

<Checkpoint>[反应4:$\mathrm{2MoS_2+7O_2\to 2MoO_3+4SO_2}$](1)</Checkpoint>

辉钼矿在氧气气氛与氢氧化钠溶液反应时硫元素可以被氧化至+6价，因此方程式为：

$\mathrm{2MoS_2+9O_2+12NaOH\to 2Na_2MoO_4+4Na_2SO_4+6H_2O}$

<Checkpoint>[反应5:$\mathrm{2MoS_2+9O_2+12NaOH\to 2Na_2MoO_4+4Na_2SO_4+6H_2O}$](1)</Checkpoint>

反应6生成含有7个钼原子同多酸盐，其方程式为：

$\mathrm{7MoO_4^{2-}+8H^++6NH_4^+\to (NH_4)_6Mo_7O_{24}+4H_2O}$

<Checkpoint>[反应6:$\mathrm{7MoO_4^{2-}+8H^++6NH_4^+\to (NH_4)_6Mo_7O_{24}+4H_2O}$](1)</Checkpoint>

因此：

||1|2|3|4|5|6|
|---|---|---|---|---|---|---|
|$a$|37|6|6|15|35|26|
|$b$|1|0|0|3|11|16|

$E(a)=20.83,D(a)=12.67,E(b)=5.17,D(b)=6.15$

$\frac{E(a)}{D(a)}=1.64,\frac{E(b)}{D(b)}=0.84$

<Checkpoint>[$\frac{E(a)}{D(a)}=1.64$](0.5)</Checkpoint>

<Checkpoint>[$\frac{E(b)}{D(b)}=0.84$](0.5)</Checkpoint>

$\log_{\frac{E(a)}{D(a)}}\frac{E(b)}{D(b)}=-0.35$