There are five chemical/ionic equations as follows:  

1. Aluminum metal, sodium azide, and ammonium chloride react at high temperature to produce aluminum nitride $\mathrm{AlN}$, with hydrogen gas as the sole gaseous product.  
2. Carbon atoms generated by low-temperature arc discharge react with excess $\mathrm{B_{2}Cl_{4}}$, with no cleavage of $\mathrm{B-Cl}$ bonds during the reaction.  
3. $\mathrm{CaAl_{2}O_{4}}$ undergoes a high-temperature hydrothermal reaction with phosphoric acid, yielding hydroxylapatite $\mathrm{Ca}_{5}(\mathrm{PO}_{4})_{3}\mathrm {OH}$ and aluminum phosphate.  
4. Tetraethoxysilane decomposes at $1023\ \mathrm{K}$ to form silicon dioxide.  
5. Assuming the carbon source is $\mathrm{HCO_3^-}$ and the nitrogen source is $\mathrm{NH_4^+}$, these react with hydrogen gas in high-pressure seawater to produce glutamic acid.  

After balancing these five equations with the simplest whole-number coefficients, for each equation, let the sum of the product coefficients be **a** and the number of product species be **b**. The number of equations among these five where both **a** and **b** are odd is:
有下列五个化学/离子方程式：
1. 铝单质，叠氮化钠和氯化铵在高温下反应制备氮化铝$\mathrm{AlN}$，产物中唯一气体为氢气。
2. 低温电弧放电产生的碳原子与过量的$\mathrm{B_{2}Cl_{4}}$发生作用，反应过程中$\mathrm{B-Cl}$键未断裂。
3. $\mathrm{CaAl_{2}O_{4}}$与磷酸高温水热反应，产物是羟基磷灰石$\mathrm{Ca}_{5}(\mathrm{PO}_{4})_{3}\mathrm {OH}$和磷酸铝。  
4. 四乙氧基硅在$1023\ \mathrm{K}$分解得到二氧化硅。  
5. 假设碳源为$\mathrm{HCO_3^-}$ ，氮源为$\mathrm{NH_4^+}$，在高压海水中与氢气反应，得到谷氨酸。

将这五个方程式以系数为最简整数比写出后，对于每个方程式，设其产物的系数和为a，产物的物种数目为b。则这五个方程中a，b均为奇数的方程数目为：
0
1
2
3
4
5
6
0
1
2
3
4
5
6
Equation 1: This is clearly a redox reaction, where sodium azide acts as a typical oxidizing agent for elemental aluminum, while itself being reduced. Since the only gaseous product is hydrogen, the formation of nitrogen gas and ammonia is not considered.  

<Checkpoint>[Formation of nitrogen gas and ammonia is not considered](0.5)</Checkpoint>  

It can be inferred that all nitrogen is converted to aluminum nitride, while chlorine does not participate in the redox reaction and only forms sodium chloride. Thus, the coefficients of ammonium chloride and sodium azide must be the same, allowing the equation to be balanced:  

<Checkpoint>[Coefficients of ammonium chloride and sodium azide are the same](0.5)</Checkpoint>  

$$ \mathrm{4Al + NaN_3 + NH_4Cl = 4AlN + NaCl + 2H_2} $$  

<Checkpoint>[$$ \mathrm{4Al + NaN_3 + NH_4Cl = 4AlN + NaCl + 2H_2} $$](1)</Checkpoint>  

Therefore, the sum of the product coefficients is 7, and the number of products is 3, both being odd.  

Equation 2: The structure of $\mathrm{B_2Cl_4}$ consists of two $\mathrm{BCl_2}$ units connected by a $\mathrm{B-B}$ bond. Since the $\mathrm{B-Cl}$ bonds do not break during the reaction, it indicates that $\mathrm{BCl_2}$ itself remains unchanged, meaning only the $\mathrm{B-B}$ bond is cleaved.  

<Checkpoint>[$\mathrm{B-B}$ bond is cleaved](0.5)</Checkpoint>  

Highly reactive carbon atoms generated by arc discharge form $\mathrm{C-B}$ bonds with it. Excess $\mathrm{B_2Cl_4}$ suggests that carbon atoms are saturated, resulting in the product $\mathrm{C(BCl_2)_4}$. The balanced equation is:  

$$ \mathrm{C + 2B_2Cl_4 = C(BCl_2)_4} $$  

<Checkpoint>[$$ \mathrm{C + 2B_2Cl_4 = C(BCl_2)_4} $$](1)</Checkpoint>  

The sum of the product coefficients is 1, and the number of products is 1, both being odd.  

Equation 3: The products are already given in the problem as $\mathrm{Ca_5(PO_4)_3OH}$ and aluminum phosphate $\mathrm{AlPO_4}$. No redox process occurs, and hydrothermal reactions typically involve the formation of water. Thus, the balanced equation is:  

$$ \mathrm{5CaAl_2O_4 + 13H_3PO_4 = Ca_5(PO_4)_3OH + 10AlPO_4 + 19H_2O} $$  

<Checkpoint>[$$ \mathrm{5CaAl_2O_4 + 13H_3PO_4 = Ca_5(PO_4)_3OH + 10AlPO_4 + 19H_2O} $$](1)</Checkpoint>  

The sum of the product coefficients is 30, and the number of products is 3, including an even number.  

Equation 4: Tetraethoxysilane contains four oxygen atoms. After forming silicon dioxide, two oxygen atoms and four ethyl groups remain. At high temperatures, oxygen atoms tend to form water, leaving behind four stable ethylene molecules. The balanced equation is:  

$$ \mathrm{Si(OC_2H_5)_4 = SiO_2 + 2H_2O + 4C_2H_4} $$  

<Checkpoint>[$$ \mathrm{Si(OC_2H_5)_4 = SiO_2 + 2H_2O + 4C_2H_4} $$](1)</Checkpoint>  

The sum of the product coefficients is 7, and the number of products is 3, both being odd.  

Equation 5: The chemical formula of glutamic acid is $\mathrm{C_5H_9NO_4}$. The left side of the equation requires five $\mathrm{HCO_3^-}$ ions and one nitrogen source, $\mathrm{NH_4^+}$. Excess oxygen atoms are balanced with hydrogen to form water. Seawater has a pH of about 8, so additional $\mathrm{NH_4^+}$ is needed to neutralize the alkali produced in the reaction. In this state, glutamic acid exists as a monovalent anion with both carboxyl protons dissociated.  

<Checkpoint>[The product glutamic acid carries one negative charge](0.5)</Checkpoint>  

Thus, the balanced equation is:  

$$ \mathrm{5HCO_3^- + 4NH_4^+ + 9H_2 = C_5H_8NO_4^- + 3NH_3 + 11H_2O} $$  

<Checkpoint>[$$ \mathrm{5HCO_3^- + 4NH_4^+ + 9H_2 = C_5H_8NO_4^- + 3NH_3 + 11H_2O} $$](1)</Checkpoint>  

The sum of the product coefficients is 15, and the number of products is 3, with no even numbers.  

In summary, among the five equations, four (1, 2, 4, and 5) have both the sum of product coefficients and the number of products as odd numbers. Therefore, the answer is E.  

<Checkpoint>[Four equations (1, 2, 4, and 5) have both sums and numbers as odd](1)</Checkpoint>
方程式1：很明显的氧化还原反应，叠氮化钠为很典型的氧化剂氧化铝单质，同时自身被还原，由于气态产物只有氢气所以不考虑氮气和氨气的生成，

<Checkpoint>[不考虑氮气和氨气的生成](0.5)</Checkpoint>

可以推断氮元素此时均被转化为氮化铝，氯元素不参与氧化还原反应只能生成氯化钠，从而氯化铵和叠氮化钠的系数相同
，即可配平方程：

<Checkpoint>[氯化铵和叠氮化钠的系数相同](0.5)</Checkpoint>

$$ \mathrm{4Al+ NaN_3 + NH_4Cl = 4AlN + NaCl+ 2H_2} $$

<Checkpoint>[$$ \mathrm{4Al+ NaN_3 + NH_4Cl = 4AlN + NaCl+ 2H_2} $$](1)</Checkpoint>

故产物系数和为7，产物数目为3，均为奇数。

方程式2：$\mathrm{B_{2}Cl_{4}}$的结构为两个$\mathrm{BCl_2}$通过$\mathrm{B-B}$键连接，反应过程中$\mathrm{B-Cl}$键未断裂说明$\mathrm{BCl_2}$本身没有发生变化，那只能是$\mathrm{B-B}$键断裂，

<Checkpoint>[$\mathrm{B-B}$键断裂](0.5)</Checkpoint>

电弧放电产生的活性很高的碳原子与其形成$\mathrm{C-B}$键。过量的$\mathrm{B_{2}Cl_{4}}$表明碳原子被饱和，从而产物明确为$\mathrm{C(BCl_2)_4}$。配平方程：

$$ \mathrm{C + 2B_2Cl_4 = C(BCl_2)_4} $$

<Checkpoint>[$$ \mathrm{C + 2B_2Cl_4 = C(BCl_2)_4} $$](1)</Checkpoint>

产物系数和为1，产物数目为1，均为奇数。

方程式3：产物在题目信息中已经给出，为$\mathrm{Ca}_{5}(\mathrm{PO}_{4})_{3}\mathrm {OH}$和磷酸铝$\mathrm{AlPO_4}$，无氧化还原过程，水热反应很容易想到还有水的生成。从而配平方程：

$$ \mathrm{5CaAl_2O_4 + 13H_3PO_4 = Ca_5(PO_4)_3OH + 10AlPO_4 + 19H_2O} $$

<Checkpoint>[$$\mathrm{5CaAl_2O_4 + 13H_3PO_4 = Ca_5(PO_4)_3OH + 10AlPO_4 + 19H_2O} $$](1)</Checkpoint>

产物系数和为30，产物数目为3，有偶数。

方程式4：四乙氧基硅含四个氧原子，在形成二氧化硅后还余两个氧原子和四个乙基，高温下氧原子形成水趋势大，刚好剩余四个稳定的乙烯小分子。从而配平方程：

$$ \mathrm{Si(OC_2H_5)_4 = SiO_2 + 2H_2O + 4C_2H_4} $$

<Checkpoint>[$$ \mathrm{Si(OC_2H_5)_4 = SiO_2 + 2H_2O + 4C_2H_4} $$](1)</Checkpoint>

产物系数和为7，产物数目为3，均为奇数。

方程式5：谷氨酸化学式为$\mathrm{C_5H_9NO_4}$，方程式左边需要五个$\mathrm{HCO_3^-}$和一个氮源$\mathrm{NH_4^+}$，多余的氧原子利用氢气配平为水。海水的pH约为8，此条件下还需多余的$\mathrm{NH_4^+}$中和反应产生的碱；谷氨酸的存在形式为一价负离子，两个羧基的质子均电离。

<Checkpoint>[产物中谷氨酸带一个负电荷](0.5)</Checkpoint>

从而配平方程式：

$$ \mathrm{5HCO_3^-+ 4NH_4^+ + 9H_2 = C_5H_8NO_4^- + 3NH_3 + 11H_2O} $$

<Checkpoint>[ $$ \mathrm{5HCO_3^-+ 4NH_4^+ + 9H_2 = C_5H_8NO_4^- + 3NH_3 + 11H_2O} $$](1)</Checkpoint>

产物系数和为15，产物数目为3，没有偶数。

综上，五个方程的产物系数和和产物数目均为奇数的有四个方程式，分别是1、2、4、5，故选E。

<Checkpoint>[均为奇数的有四个方程式，分别是1、2、4、5](1)</Checkpoint>