For 2 SO2(g) + O2(g) equilibrium reaction arrow 2SO3(g), Kp = 3.0 104 at 700. K. In a 2.00-L vessel the equilibrium mixture contains 1.11 g of SO3 and 0.106 g of O2. How many grams of SO2 are in the vessel

See your later post above.

sdafn

To determine the number of grams of SO2 in the vessel, we need to use the given information and the equilibrium constant (Kp) for the reaction.

First, let's convert the given masses of SO3 and O2 to the number of moles. We can use the molar masses of the compounds:

Molar mass of SO3: 32.06 g/mol
Molar mass of O2: 32.00 g/mol

Number of moles of SO3 = mass / molar mass = 1.11 g / 32.06 g/mol
Number of moles of O2 = mass / molar mass = 0.106 g / 32.00 g/mol

Now, we need to calculate the number of moles of SO2 using the stoichiometry of the balanced equation. According to the equation, 2 moles of SO2 are consumed for every 1 mole of O2. So, the number of moles of SO2 can be calculated as follows:

Number of moles of SO2 = 2 * (mole of O2)

Next, we need to determine the concentration of each component (SO2, SO3, O2) in the vessel by dividing the number of moles by the volume of the vessel:

Concentration of SO3 = (mole of SO3) / (volume of vessel in L)
Concentration of O2 = (mole of O2) / (volume of vessel in L)
Concentration of SO2 = (mole of SO2) / (volume of vessel in L)

Finally, we can use the equilibrium constant (Kp) expression to calculate the number of moles of SO2 at equilibrium:

Kp = (concentration of SO3)^2 * (concentration of O2) / (concentration of SO2)^2

Substituting the given values and rearranging the equation, we can solve for the concentration of SO2. Since we know the concentration of SO2, we can also find the number of moles of SO2. Then, by multiplying the number of moles by the molar mass of SO2, we can calculate the mass of SO2 in grams.

Note: The value of Kp and the temperature used in this calculation are specific to the given equilibrium reaction.

It is important to note that this process assumes the reaction has reached equilibrium and that there are no additional factors affecting the reaction.