Silver sulfate decomposes at high temperatures, producing a solid and multiple gases. The density of the gaseous products (treated as ideal gases) converted to standard conditions is 2.57 g/L. Based on this, the equation for the reaction can be determined. When the coefficients are reduced to the simplest integer ratio, let the coefficient of the solid product be A and the sum of the coefficients of all gases be B. Provide the values of A and B, respectively.
硫酸银在高温下分解，生成一种固体和多种气体。生成物中气相的密度（视为理想气体）折算至标况下为2.57 g/L，据此可以得出该反应的方程式。在系数约至最简整数比的情况下，令生成物中固体的系数为A，所有气体系数之和为B，分别给出A和B的值。
A=6, B=5
A=3, B=5
A=5, B=6
A=4, B=4
A=5, B=5
A=4, B=3
A=2, B=3
A=4, B=5
A=6, B=5
A=3, B=5
A=5, B=6
A=4, B=4
A=5, B=5
A=4, B=3
A=2, B=3
A=4, B=5
At high temperatures, $Ag_2O$ will decompose, and $SO_3$ may decompose. Therefore, the solid product is $Ag$, and the gaseous products are $SO_3$, $SO_2$, and $O_2$.

<Checkpoint>[The solid product is $Ag$](1)</Checkpoint>

<Checkpoint>[The gaseous products are $SO_3$, $SO_2$, and $O_2$](2)</Checkpoint>

The reaction can be written as:

$Ag_2SO_4 = 2 Ag + a SO_3 + b SO_2 + c O_2 $

<Checkpoint>[$Ag_2SO_4 = 2 Ag + a SO_3 + b SO_2 + c O_2 $](1)</Checkpoint>

According to $M = \rho \cdot 22.4 (g/mol)$, the average molecular weight of the gaseous products is calculated as $M_{avg} = 57.6 g/mol$.

<Checkpoint>[$M_{avg} = 57.6 g/mol$](1)</Checkpoint>

(Verification shows that generating only $SO_3$ + $O_2$ or $SO_2$ + $O_2$ would contradict this.)

Based on the conservation of S atoms, $a+b = 1$.

<Checkpoint>[$a+b = 1$](1)</Checkpoint>

Based on the conservation of O atoms, $3a+2b+2c = 4$.

<Checkpoint>[$3a+2b+2c = 4$](1)</Checkpoint>

Thus, the total moles of gas $n_{gas} = a + b + c = c +1$.

<Checkpoint>[$n_{gas} = c +1$](1)</Checkpoint>

If 1 mol of the substrate decomposes, the total mass of the gas obtained is the molar mass of "$SO_4$," so:

$M_{avg} = m_{SO_4}/n_{gas} = 96/(c+1) $

<Checkpoint>[$M_{avg}  = 96/(c+1) $](2)</Checkpoint>

Substituting the data yields $c = 2/3, a = 2/3, b = 1/3$.

<Checkpoint>[$c = 2/3$](1)</Checkpoint>

<Checkpoint>[$a = 2/3$](0.5)</Checkpoint>

<Checkpoint>[$b = 1/3$](0.5)</Checkpoint>

By simplifying the coefficients of the equation to the smallest integer ratio, the actual equation is:

$3Ag_2SO_4 = 6 Ag + 2 SO_3 + SO_2 + 2 O_2 $

<Checkpoint>[$3Ag_2SO_4 = 6 Ag + 2 SO_3 + SO_2 + 2 O_2 $](1)</Checkpoint>

Therefore, $A=6, B=2+1+2=5$.

<Checkpoint>[$A=6$](0.5)</Checkpoint>

<Checkpoint>[$B=5$](0.5)</Checkpoint>
高温下$Ag_2O$将分解，$SO_3$可能分解，因此固体产物是$Ag$，气体产物是$SO_3$、$SO_2$、$O_2$。

<Checkpoint>[固体产物是$Ag$](1)</Checkpoint>

<Checkpoint>[气体产物是$SO_3$、$SO_2$、$O_2$](2)</Checkpoint>


反应式可以写成

$Ag_2SO_4 = 2 Ag + a SO_3 + b SO_2 + c O_2 $

<Checkpoint>[$Ag_2SO_4 = 2 Ag + a SO_3 + b SO_2 + c O_2 $](1)</Checkpoint>

根据$M = \rho \cdot 22.4 (g/mol)$，算得气体产物的平均分子量$M_{avg} = 57.6 g/mol$。

<Checkpoint>[$M_{avg} = 57.6 g/mol$](1)</Checkpoint>

（经验证可发现只生成$SO_3$ + $O_2$或$SO_2$ + $O_2$将与此矛盾）

根据S原子数守恒，$a+b = 1$。

<Checkpoint>[$a+b = 1$](1)</Checkpoint>

根据O原子数守恒，$3a+2b+2c = 4$。

<Checkpoint>[$3a+2b+2c = 4$](1)</Checkpoint>

因此气体的总摩尔数$n_{gas} = a + b + c = c +1$。

<Checkpoint>[$n_{gas} = c +1$](1)</Checkpoint>

若分解1 mol底物，得到的气体总质量为“$SO_4$”的摩尔质量，因此

$M_{avg} = m_{SO_4}/n_{gas} = 96/(c+1) $

<Checkpoint>[$M_{avg}  = 96/(c+1) $](2)</Checkpoint>

代入数据解得 $c = 2/3,a = 2/3,b = 1/3$。

<Checkpoint>[$c = 2/3$](1)</Checkpoint>

<Checkpoint>[$a = 2/3$](0.5)</Checkpoint>

<Checkpoint>[$b = 1/3$](0.5)</Checkpoint>

将方程式系数化为最简整数比，实际的方程式为

$3Ag_2SO_4 = 6 Ag + 2 SO_3 + SO_2 + 2 O_2 $

<Checkpoint>[$3Ag_2SO_4 = 6 Ag + 2 SO_3 + SO_2 + 2 O_2 $](1)</Checkpoint>

因此$A=6, B=2+1+2=5$。

<Checkpoint>[$A=6$](0.5)</Checkpoint>

<Checkpoint>[$B=5$](0.5)</Checkpoint>