In liquid ammonia at $\mathrm{238\ K}$, sodium and white phosphorus react at a certain molar ratio to form a yellow crystal. Further analysis indicates that this yellow crystal is a $1:1$ ionic compound; both its cation and anion are mononuclear sodium coordination ions, each with a coordination number of $5$. In this reaction, white phosphorus is converted solely into the phosphorus-containing ligand in the anion of the ionic compound. The molar mass of this ligand is $\mathrm{96\ g/mol}$, and all phosphorus atoms satisfy the octet rule.  

In the chemical equation for the formation of this yellow crystal, what is the sum of the coefficients of all substances on both sides of the equation (using the simplest integer ratio for the coefficients)?
在$\mathrm{238\ K}$下的液氨中,钠与白磷按一定的摩尔比混合反应,生成一种黄色晶体。进一步分析表明,该黄色晶体为$1:1$型离子化合物;其阴阳离子中均为$\mathrm{Na}$的单核配离子,配位原子数均为$5$。该反应中,白磷仅转化为了该离子化合物阴离子中的含磷配体;该配体摩尔质量为$\mathrm{96\ g/mol}$,所有磷原子均满足8电子结构。

在生成该黄色晶体的化学方程式中，方程式左右两端各物质的系数之和是多少（系数为最简整数比）？
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None of the above
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以上都不是
In a system containing only the four elements $\mathrm{Na, P, H, N}$, the phosphorus-containing ligand in the anion should consist solely of $\mathrm{P, H, N}$ elements. Based on molar mass, the possible chemical formulas for the ligand are $\mathrm{PN_4H_9^-, P_2N_2H_6, or P_3H_3^{2-}}$. Considering the octet rule for phosphorus, the ligand cannot be $\mathrm{PN_4H_9^-}$. If the ligand were $\mathrm{P_2N_2H_6}$, the complex formed with sodium ions could not carry a negative charge. Therefore, the chemical formula of the ligand must be $\mathrm{P_3H_3^{2-}}$.  

<Checkpoint>[The ligand's chemical formula is $\mathrm{P_3H_3^{2-}}$](1)</Checkpoint>  

Since the cation is five-coordinate, the cation is $\mathrm{Na(NH_3)_5^+}$.  

<Checkpoint>[The cation is $\mathrm{Na(NH_3)_5^+}$](0.5)</Checkpoint>  

Given that the anion is five-coordinate and the $\mathrm{P_3H_3^{2-}}$ ligand in the anion is chelating, the remaining three coordination sites are occupied by $\mathrm{NH_3}$, making the anion $\mathrm{Na(P_3H_3)(NH_3)_3^-}$.  

<Checkpoint>[The anion is $\mathrm{Na(P_3H_3)(NH_3)_3^-}$](1)</Checkpoint>  

Thus, the chemical formula of the yellow crystal is $\mathrm{[Na(NH_3)_5][Na(P_3H_3)(NH_3)_3]}$. The chemical formula of white phosphorus is $\mathrm{P_4}$.  

<Checkpoint>[The chemical formula of the yellow crystal is $\mathrm{[Na(NH_3)_5][Na(P_3H_3)(NH_3)_3]}$](0.5)</Checkpoint>  

The left side of the equation consists of $\mathrm{Na, P_4, NH_3}$, while the right side includes $\mathrm{[Na(NH_3)_5][Na(P_3H_3)(NH_3)_3]}$ and $\mathrm{NaNH_2}$. Balancing the equation yields:  
$\mathrm{20Na + 3P_4 + 44NH_3 \to 4[Na(NH_3)_5][Na(P_3H_3)(NH_3)_3] + 12NaNH_2}$  

<Checkpoint>[The equation is $\mathrm{20Na + 3P_4 + 44NH_3 \to 4[Na(NH_3)_5][Na(P_3H_3)(NH_3)_3] + 12NaNH_2}$](1)</Checkpoint>  

The sum of the coefficients is 83, so option T is correct.  

<Checkpoint>[The sum of coefficients is 83](0.5)</Checkpoint>
在体系中仅含$\mathrm{Na, P, H, N}$四种元素，则阴离子中的含磷配体应仅含$\mathrm{P, H, N}$元素。由摩尔质量可知，该配体化学式可能为$\mathrm{PN_4H_9^-, P_2N_2H_6或P_3H_3^{2-}}$。由磷元素满足八隅体规则可知，配体不为$\mathrm{PN_4H_9^-}$；若配体为$\mathrm{P_2N_2H_6}$，则钠离子与该配体形成的配合物不可能带负电荷。因此配体化学式只能为$\mathrm{P_3H_3^{2-}}$。

<Checkpoint>[配体化学式为$\mathrm{P_3H_3^{2-}}$](1)</Checkpoint>

由阳离子为五配位，可知阳离子为$\mathrm{Na(NH_3)_5^+}$。

<Checkpoint>[阳离子为$\mathrm{Na(NH_3)_5^+}$](0.5)</Checkpoint>

由阴离子为五配位，且阴离子中$\mathrm{P_3H_3^{2-}}$配体为螯合配位，可知剩下的三个配位数由$\mathrm{NH_3}$提供，阴离子为$\mathrm{Na(P_3H_3)(NH_3)_3^-}$。

<Checkpoint>[阴离子为$\mathrm{Na(P_3H_3)(NH_3)_3^-}$](1)</Checkpoint>

则黄色晶体的化学式为$\mathrm{[Na(NH_3)_5][Na(P_3H_3)(NH_3)_3]}$。而白磷化学式为$\mathrm{P_4}$。

<Checkpoint>[黄色晶体的化学式为$\mathrm{[Na(NH_3)_5][Na(P_3H_3)(NH_3)_3]}$](0.5)</Checkpoint>

则方程式左侧为$\mathrm{Na，P_4，NH_3}$，右侧为$\mathrm{[Na(NH_3)_5][Na(P_3H_3)(NH_3)_3]}$和$\mathrm{NaNH_2}$。配平可得方程式为
$\mathrm{20Na + 3P_4 + 44NH_3 \to 4[Na(NH_3)_5][Na(P_3H_3)(NH_3)_3] + 12NaNH_2}$

<Checkpoint>[方程式为 $\mathrm{20Na + 3P_4 + 44NH_3 \to 4[Na(NH_3)_5][Na(P_3H_3)(NH_3)_3] + 12NaNH_2}$](1)</Checkpoint>

则系数之和为83，选项T正确。

<Checkpoint>[系数之和为83](0.5)</Checkpoint>