$\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ represent three common elements, while $\mathbf{M}$ and $\mathbf{N}$ are two compounds that can be considered the same type of substance under the classification of "acid, base, salt, oxide, or elemental substance." They conform to the following chemical reaction processes:
$$
A_2B+ACB_3 = M
$$
$$
A+ACB_3=M+C_2
$$
$$
ACB_2+A_2B_2=M
$$
$$
A+ACB_2=N+C_2
$$
These equations are not balanced. Please provide the chemical formulas of $\mathbf{M}$ and $\mathbf{N}$.
$\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$分别代表三种常见元素，$\mathbf{M}$与$\mathbf{N}$为两种化合物，可以认为在“酸、碱、盐、氧化物、单质”的分类下是同类物质。他们符合以下化学反应过程。
$$
A_2B+ACB_3 = M
$$
$$
A+ACB_3=M+C_2
$$
$$
ACB_2+A_2B_2=M
$$
$$
A+ACB_2=N+C_2
$$
这些方程式并未配平，请给出$\mathbf{M}$, $\mathbf{N}$的化学式。
M: $\mathrm{H_2CO_3}$,
N: $\mathrm{CO_2}$
M: $\mathrm{Na_2CO_3}$,
N: $\mathrm{NaHCO_3}$
M: $\mathrm{Na_3NO_4}$,
N: $\mathrm{Na_3NO_3}$
M: $\mathrm{Na_2SO_4}$,
N: $\mathrm{Na_2SO_3}$
M: $\mathrm{H_2SO_4}$,
N: $\mathrm{SO_2}$
M: $\mathrm{CO_2}$,
N: $\mathrm{CO}$
M: $\mathrm{Na_3SO_4}$,  
N: $\mathrm{Na_3SO_3}$
M: $\mathrm{Na_2S_2O_4}$,  
N: $\mathrm{Na_2SO_3}$
M: $\mathrm{Na_2N_2O_4}$,  
N: $\mathrm{Na_2NO_3}$
M: $\mathrm{Na_2CO_3}$,  
N: $\mathrm{Na_2C_2O_4}$
M: $\mathrm{H_2CO_3}$,
N: $\mathrm{CO_2}$
M: $\mathrm{Na_2CO_3}$,
N: $\mathrm{NaHCO_3}$
M: $\mathrm{Na_3NO_4}$,
N: $\mathrm{Na_3NO_3}$
M: $\mathrm{Na_2SO_4}$,
N: $\mathrm{Na_2SO_3}$
M: $\mathrm{H_2SO_4}$,
N: $\mathrm{SO_2}$
M: $\mathrm{CO_2}$,
N: $\mathrm{CO}$
M: $\mathrm{Na_3SO_4}$,
N: $\mathrm{Na_3SO_3}$
M: $\mathrm{Na_2S_2O_4}$,
N: $\mathrm{Na_2SO_3}$
M: $\mathrm{Na_2N_2O_4}$,
N: $\mathrm{Na_2NO_3}$
M: $\mathrm{Na_2CO_3}$,
N: $\mathrm{Na_2C_2O_4}$
First, the obtained chemical formulas should be reasonable existing forms. Option G contains $\mathrm{Na_3SO_4}$, which does not exist, and Option I contains $\mathrm{Na_2NO_3}$, which does not exist. Therefore, Options G and I are incorrect.

<Checkpoint>[Option G contains $\mathrm{Na_3SO_4}$, which does not exist, and Option I contains $\mathrm{Na_2NO_3}$, which does not exist. Therefore, Options G and I are incorrect.](1)</Checkpoint>

Given that $\mathbf{M}$ and $\mathbf{N}$ in the question are two compounds, they can be considered as the same type of substances under the classification of "acid, base, salt, oxide, or simple substance." Thus, Options A and E can be excluded.

<Checkpoint>[In Options A and E, M and N belong to acid and oxide, respectively, and are not the same type of substances. Therefore, Options A and E are excluded.](1)</Checkpoint>

According to the addition reaction equation, M must contain all three elements A, B, and C, and the number of atoms of A and B in M must be greater than that of C. Option F contains only two elements and is incorrect. Option H does not conform to the atomic count rule and is also incorrect.

<Checkpoint>[According to the addition reaction equation, M must contain all three elements A, B, and C, and the number of atoms of A and B in M must be greater than that of C. Option F contains only two elements and is incorrect. Option H does not conform to the atomic count rule and is also incorrect.](1)</Checkpoint>

$$
A_2B+ACB_3 = M
$$

$$
ACB_2+A_2B_2=M
$$

If the coefficients of the reactants are all 1, M can be derived as $A_3CB_4$. Among the options, Option C corresponds to this formula, which, although uncommon, exists. Substituting $\mathbf{A}$=$\mathrm{Na}$, $\mathbf{B}$=$\mathrm{O}$, and $\mathbf{C}$=$\mathrm{N}$ into the remaining equations:

$$
A+ACB_2=N+C_2
$$
$$
A+ACB_3=M+C_2
$$

From the equations, N can contain at most three elements, so Option B is incorrect.
<Checkpoint>[N can contain at most three elements, so Option B is incorrect.](0.5)</Checkpoint>

The generated N: $\mathrm{Na_3NO_3}$ is uncommon but exists, and the production of $N_2$ is reasonable. The other chemical formulas are also reasonable, so Option C is correct.
Specifically: M: $\mathrm{Na_3NO_4}$, N: $\mathrm{Na_3NO_3}$

<Checkpoint>[M: $\mathrm{Na_3NO_4}$, N: $\mathrm{Na_3NO_3}$](1)</Checkpoint>

For Options D and J, A is $Na$, B is $S$ or $C$, and C is $O$. Substituting these into the chemical formulas reveals the existence of unreasonable substances such as $NaCO_2$ and $NaSO_3$. Therefore, Options D and J are incorrect.

<Checkpoint>[When A is $Na$, B is $S$ or $C$, and C is $O$, the equations contain unreasonable substances such as $NaCO_2$ and $NaSO_3$. Therefore, Options D and J are incorrect.](1)</Checkpoint>
首先，得到的化学式应当是合理的存在形式。选项G中$\mathrm{Na_3SO_4}$不存在，选项I中 $\mathrm{Na_2NO_3}$不存在，因此选项G、I错误。

<Checkpoint>[选项G中$\mathrm{Na_3SO_4}$不存在，选项I中 $\mathrm{Na_2NO_3}$不存在，因此选项G、I错误。](1)</Checkpoint>

根据题目中$\mathbf{M}$与$\mathbf{N}$为两种化合物，可以认为在“酸、碱、盐、氧化物、单质”的分类下是同类物质，可以排除选项A和选项E。

<Checkpoint>[选项A和选项E中的M、N分别属于酸和氧化物，不是同类物质，排除选项A和选项E。](1)</Checkpoint>

根据加成反应方程式，M一定含有A、B、C三种元素，且M中含有A和B的原子个数一定大于C，选项F只含有两种元素，错误。选项H原子个数规律不符合，错误。

<Checkpoint>[根据加成反应方程式，M一定含有A、B、C三种元素，且M中含有A和B的原子个数一定大于C，选项F只含有两种元素，错误。选项H原子个数规律不符合，错误](1)</Checkpoint>



$$
A_2B+ACB_3 = M
$$

$$
ACB_2+A_2B_2=M
$$

若反应物系数均为1，可以得到$M$为$A_3CB_4$。对应选项里C，其虽然不常见但存在。对应$\mathbf{A}$=$\mathrm{Na}$，$\mathbf{B}$=$\mathrm{O}$，$\mathbf{C}$=$\mathrm{N}$。带入剩余式子：
 
$$
A+ACB_2=N+C_2
$$
$$
A+ACB_3=M+C_2
$$

通过方程式N最多含有三种元素，选项B错误。
<Checkpoint>[N最多含有三种元素，选项B错误](0.5)</Checkpoint>

生成的N：$\mathrm{Na_3NO_3}$不常见但存在，产生$N_2$合理，其余化学式也都合理，选项C正确。
即：M: $\mathrm{Na_3NO_4}$，N: $\mathrm{Na_3NO_3}$

<Checkpoint>[M: $\mathrm{Na_3NO_4}$, N: $\mathrm{Na_3NO_3}$](1)</Checkpoint>

对于D选项和J选项，A为$Na$，B为$S$
或$C$，C为$O$。带入各化学式分别存在$NaCO_2$、$NaSO_3$的不合理物质，因此选项D和J错误。

<Checkpoint>[A为$Na$，B为$S$
或$C$，C为$O$时，方程式中有$NaCO_2$、$NaSO_3$的不合理物质，因此选项D和J错误](1)</Checkpoint>