At 1173 k, keq = 0.0108 for the following reaction: caco3(s) ⇄ cao(s) + co2(g) the reaction takes place in a 10.0 l vessel at 1173 k. if a mixture of 15.0 g caco3, 15.0 g cao, and 4.25 g co2 is allowed to approach equilibrium, what will happen to the amount of caco3? group of answer choices
- it will remain the same
- it will increase
- not enough information is provided to answer this question
- it will decrease

The amount of calcium carbonate will increase.

Explanation:

K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

K>Q , reaction will move forward by making more product.K<Q , reaction will move backward by making more reactant.

Equilibrium constant of the reaction =

Concentration of

Concentration of

Concentration of

This means that equilibrium will move in backward direction by which amount of calcium carbonate will increase.

Because it was a dramatically big guy

If it releases 5 tons a day and there are 7 days a take that answer and multiply by 5 percent (0.05) to find the amount released in one week. leave a comment if you need more .