Iron powder reacts with nitric acid of different concentrations at various temperatures, generating multiple nitrogen-containing reduction products, including one salt and several gases. A series of gaseous products obtained from the reaction of iron with excess nitric acid at different temperatures and concentrations were collected. After drying, their density was measured to be $1.545 \times 10^{-3} g/cm^3$ under standard conditions. After passing these gases through a $NaOH$ solution, the density changed to $1.416 \times 10^{-3} g/cm^3$. Subsequently, after passing through a $KMnO_4$ solution, the density became $1.429 \times 10^{-3} g/cm^3$. (All densities are measured under standard conditions.)

It is known that after each absorption step by the solution, the gases were dried appropriately to remove water vapor without additional gas loss.

Considering only the gaseous products generated from the reaction between iron and nitric acid, determine the coefficient of nitric acid in the overall chemical reaction equation when simplified to the smallest whole number ratio. (Assume that when the ratios of all gases are simplified to the smallest whole numbers, their coefficients are single-digit.)
铁粉在不同温度下，与不同浓度的硝酸反应，会生成多种含$N$的还原产物，包括一种盐和多种气体。现收集了一系列铁与过量不同温度、不同浓度的硝酸反应得到的气体产物，经过干燥后它们的密度为$1.545 \times 10^{-3} g/cm^3$。将这些气体通过$NaOH$溶液后，密度变为$1.416 \times 10^{-3} g/cm^3$。再通过$KMnO_4$溶液后，密度变为$1.429 \times 10^{-3} g/cm^3$。（密度均为标准状况下的密度）

已知以上每一步通过溶液吸收后都进行了干燥，每一步干燥都选择了合适的方法除去水汽，没有额外的气体损耗。

若只考虑铁与硝酸反应生成气体产物的情况，求将总的化学反应方程式化为最简整数比后，反应物中硝酸的系数。（已知所有气体的比例化为最简整数比时，系数都是个位数）
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182
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182
The reaction of iron with nitric acid can produce possible products such as $NH_4NO_3, N_2, NO, N_2O, NO_2$. In the problem, the step involving $NaOH$ solution is where $NO$ and $NO_2$ are absorbed by $NaOH$, undergoing a comproportionation reaction. Additionally, if $NO_2$ is in excess compared to $NO$, the excess $NO_2$ can also directly react with the $NaOH$ solution and be completely consumed. Therefore, the mixed gas after passing through the $NaOH$ solution cannot contain $NO_2$. The change in gas density before and after passing through the $KMnO_4$ solution indicates that the mixed gas before passing through the $KMnO_4$ solution still contains $NO$, and the oxidation process of $NO$ occurs.

<Checkpoint>[The product gases include $N_2, N_2O, NO, NO_2$](1)</Checkpoint>

The density of $1.545 \times 10^{-3} g/cm^3$ under standard conditions corresponds to a molecular weight of $1.545 \times 10^{-3} \times 22.413 \times 10^3 = 34.63 g/mol$; the density of $1.416 \times 10^{-3} g/cm^3$ corresponds to a molecular weight of $1.416 \times 10^{-3} \times 22.413 \times 10^3 = 31.74 g/mol$; the density of $1.429 \times 10^{-3} g/cm^3$ corresponds to a molecular weight of $1.429 \times 10^{-3} \times 22.413 \times 10^3 = 32.03 g/mol$.

<Checkpoint>[The three densities correspond to molecular weights of $34.62, 31.74, 32.03 g/mol$](1)</Checkpoint>

Assuming the proportion of $N_2$ is 1, let the ratio of $N_2O$ to $N_2$ be x. Then the reduced mass of the two can be expressed as:

$[1 \times 14.01 \times 2 + x \times (14.01 \times 2 + 16.00)] / (1 + x) = 32.03$

Solving for x gives $x = 0.334 \approx 1/3$

Therefore, $N_2:N_2O = 3:1$

Assuming after passing through the $NaOH$ solution, $NO:N_2 = y$, then the reduced mass of $NO, N_2, N_2O$ can be expressed as:

$[3 \times 14.01 \times 2 + 1 \times (14.01 \times 2 + 16.00) + y \times (14.01 + 16.00)] / (3 + 1 + y) = 31.74$

Solving for y gives $y = 0.64 \approx 2/3$ (the denominator cannot exceed 3, otherwise the coefficient of $N_2$ would exceed 10)

Assuming $NO_2:N_2O = z$, then the reduced mass of $NO_2, NO, N_2, N_2O$ can be expressed as:

$[3 \times 14.01 \times 2 + 1 \times (14.01 \times 2 + 16.00) + (2/3 + z) \times (14.01 + 16.00) + z \times (14.01 + 16.00 \times 2)] / (3 + 1 + 2/3 + 2z) = 34.63$

Solving for z gives $z = 1.997 \approx 2$

Therefore, $N_2:N_2O:NO:NO_2 = 3:1:(2/3 + 2):2 = 9:3:8:6$

<Checkpoint>[$N_2:N_2O:NO:NO_2 = 9:3:8:6$](2)</Checkpoint>

Balancing the equation:

$48Fe + 182HNO_3 = 48Fe(NO_3)_3 + 9N_2 + 3N_2O + 8NO + 6NO_2 + 91H_2O$

Alternatively, the coefficients can be calculated directly based on the charge required for the redox reaction:

$9 \times 10 + 3 \times 8 + 8 \times 3 + 6 \times 1 = 144$

The coefficient for $Fe$ is $144/3 = 48$

The total number of $N$ is $9 \times 2 + 3 \times 2 + 8 + 6 + 48 \times 3 = 182$

Therefore, the coefficient is 182.

<Checkpoint>[The coefficient of $HNO_3$ is 182](2)</Checkpoint>
铁与硝酸反应，可能的产物有$NH_4NO_3, N_2, NO, N_2O, NO_2$。题目中通过$NaOH$溶液一步是$NO$与$NO_2$被$NaOH$吸收，发生归中反应，同时若$NO_2$相较$NO$过量、过量的$NO_2$也能直接和$NaOH$溶液发生反应被消耗完，因此通过$NaOH$溶液后混合气体不可能含有$NO_2$。通过$KMnO_4$溶液前后气体密度发生变化，说明通过$KMnO_4$溶液前混合气体中还含有$NO$，发生了$NO$的氧化过程。

<Checkpoint>[产物气体有$N_2, N_2O, NO, NO_2$](1)</Checkpoint>

标准状况下密度$1.545 \times 10^{-3} g/cm^3$对应分子量$1.545 \times 10^{-3} \times 22.413 \times 10^3 = 34.63 g/mol$；密度$1.416 \times 10^{-3} g/cm^3$对应分子量$1.416 \times 10^{-3} \times 22.413 \times 10^3 = 31.74 g/mol$；密度$1.429 \times 10^{-3} g/cm^3$对应分子量$1.429 \times 10^{-3} \times 22.413 \times 10^3 = 32.03 g/mol$。

<Checkpoint>[3种密度对应分子量分别为$34.62, 31.74, 32.03 g/mol$](1)</Checkpoint>

以$N_2$的比例为1，设$N_2O$与$N_2$的比值为x，则两者的折合质量可以表示为：

$[1 \times 14.01 \times 2+x \times (14.01 \times 2+16.00)]/(1+x)=32.03$

解得$x = 0.334 \approx 1/3$

因此$N_2:N_2O=3:1$

设通过$NaOH$溶液后，$NO:N_2=y$，则$NO, N_2, N_2O$的折合质量可以表示为：

$[3 \times 14.01 \times 2+1 \times (14.01 \times 2+16.00) + y \times (14.01+16.00)]/(3+1+y)=31.74$

解得$y = 0.64 \approx 2/3$（分母最大为3，否则$N_2$的系数超过10）

设$NO_2:N_2O=z$，则$NO_2, NO, N_2, N_2O$的折合质量可以表示为：

$[3 \times 14.01 \times 2+1 \times (14.01 \times 2+16.00) + (2/3+z) \times (14.01+16.00) + z \times (14.01+16.00 \times 2) ]/(3+1+2/3+2z)=34.63$

解得$z = 1.997 \approx 2$

因此$N_2:N_2O:NO:NO_2=3:1:(2/3+2):2=9:3:8:6$

<Checkpoint>[$N_2:N_2O:NO:NO_2=9:3:8:6$](2)</Checkpoint>

配平方程式：

$48Fe + 182HNO_3 = 48Fe(NO_3)_3 + 9N_2 + 3N_2O + 8NO + 6NO_2 + 91H_2O$

也可根据氧化还原反应所需电荷量直接计算：

$9 \times 10 + 3 \times 8 + 8 \times 3 + 6 \times 1 = 144$

对应$Fe$的系数为$144/3=48$

则$N$的总数为$9 \times 2 + 3 \times 2 + 8 + 6 + 48 \times 3 = 182$

因此系数是182。

<Checkpoint>[$HNO_3$的系数为182](2)</Checkpoint>