A water bottling company is looking to achieve maximum production from the following reversible reaction. If the reaction has reached a point of equilibrium, what would be the best way to maximize production of water? 2 H2+O2⇔2 H2O Remove hydrogen and allow the reaction to reach equilibrium again. Stop of the process because no more water can be made after equilibrium is reached. Remove the already produced water and allow the reaction to reach equilibrium again. Remove oxygen and allow the reaction to reach equilibrium again.

Remove the already produced water and allow the reaction to reach equilibrium again.

Explanation:

According to Le Chatelier principle, when a reaction is in equilibrium and one of the factors that influence the rate of reaction is altered, the equilibrium will shift so as to annul the effects of the change.

If the product is continuously being removed from a reaction that is in equilibrium, more product will continued to be formed in another to annul the effect of reduction in the concentration of product.

Hence, in order to maximize production of water in the reaction, the product (water) needs to be removed and the reaction allowed to reach equilibrium again.

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.