Answer :
Using Henry's Law 1.50 g of [tex]CO_2[/tex] would escape from the solution when the partial pressure is reduced from 8.00 atm to 4.00 atm, assuming the solubility of [tex]CO_2[/tex] would be halved.
The problem presented is primarily a matter of applying Henry's Law, which relates the solubility of a gas in a liquid to the partial pressure of that gas above the liquid. The law is usually expressed in the form S = kH * P, where S is the solubility (concentration) of the gas, kH is Henry's Law constant, and P is the partial pressure of the gas.
In this case, a solution initially saturated with [tex]CO_2[/tex] at 8.00 atm has 3 grams of [tex]CO_2[/tex]. When the pressure is halved to 4.00 atm, Henry's Law suggests that the solubility of [tex]CO_2[/tex] would also be halved (assuming that the Henry's Law constant remains unchanged at this lower pressure).
Since the initial amount is 3.00 g at 8.00 atm, at 4.00 atm the solubility would be 1.50 g. Therefore, the mass of [tex]CO_2[/tex] that would escape is the difference: 3.00 g - 1.50 g = 1.50 g.