Answer:
b. The amount of SO3(g) decreases and the value for K increases.
Explanation:
Hello,
In this case, for the given reaction:

The change in the stoichiometric coefficient is:

In such a way, since the reagents have more moles than the products, based on Le Chatelier's principle, if the volume is increased, the side with more moles is favored. In addition, since the formation of reagent is favored, K is diminished based on the law of mass action shown below:
![K=\frac{[SO_3]_{eq}^2}{[SO_2]_{eq}^2[O_2]_{eq}}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BSO_3%5D_%7Beq%7D%5E2%7D%7B%5BSO_2%5D_%7Beq%7D%5E2%5BO_2%5D_%7Beq%7D%7D)
Therefore the answer is:
b. The amount of SO3(g) decreases and the value for K increases.
Best regards.
Answer:
it's the 2 I'm pretty sure
Answer:
2:3
You'd need 33.3 moles of potassium chlorate, KClO3 , to produce that much oxygen. Notice that you have a 2:3. mole ratio between potassium chlorate and oxygen gas, which means that, regardless of how many moles of the former react, you'll always produce 3/2 times more moles of the latter.
Explanation: