Before it is released it as potential energy and after it has been released it transforms into kinetic energy.
At equivalence there is no more HA and no more NaOH, for this particular reaction. So that means we have a beaker of NaA and H2O. The H2O contributes 1 x 10-7 M hydrogen ion and hydroxide ion. But NaA is completely soluble because group 1 ion compounds are always soluble. So NaA breaks apart in water and it just so happens to be in water. So now NaA is broken up. The Na+ doesn't change the pH but the A- does change the pH. Remember that the A anion is from a weak acid. That means it will easily attract a hydrogen ion if one is available. What do you know? The A anion is in a beaker of H+ ions! So the A- will attract H+ and become HA. When this happens, it leaves OH-, creating a basic solution, as shown below.
Answer:
1. 4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2
2. 6 moles of Cl2
Explanation:
1. The balanced equation for the reaction. This is illustrated below:
4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2
2. Determination of the number of mole of Cl2 produce when 4 moles of FeCl3 react with 4 moles. To obtain the number of mole of Cl2 produced, we must determine which reactant is the limiting reactant.
This is illustrated below:
From the balanced equation above,
4 moles of FeCl3 reacted with 3 moles of O2.
Since lesser amount of O2 (i.e 3 moles) than what was given (i.e 4 moles) is needed to react completely with 4 moles of FeCl3, therefore FeCl3 is the limiting reactant and O2 is the excess reactant.
Finally, we can obtain the number of mole Cl2 produced from the reaction as follow:
Note: the limiting reactant is used as it will produce the maximum yield of the reaction since all of it is used up in the reaction.
From the balanced equation above,
4 moles of FeCl3 will react to produced 6 moles of Cl2.
