A) At 800 K the equilibrium constant for I2(g) <=> 2 I(g) is Kc = 3.1 10-5. If an equilibrium mixture in a 10.0-L vessel contains 3.44*10-2 g of I(g), how many grams of I2 are in the mixture? answer is .060g

B)For 2 SO2(g) + O2(g)<=> 2 SO3(g), Kp = 3.0*104 at 700 K. In a 2.00-L vessel the equilibrium mixture contains 1.03 g of SO3 and 0.170 g of O2. How many grams of SO2 are in the vessel?

I don't know how to do B

On #2:

Convert 1.03 g SO3 to moles, then to pressure using PV = nRT.
Do the same for 0.170g O2.
Substitute into Kp expression and solve for pSO2. Finally, use PV = nRT to solve for moles and convert to grams.

To solve problem B, you need to use the given equilibrium constant expression and the given amounts of SO3 and O2 in the equilibrium mixture to find the amount of SO2.

Step 1: Write the balanced chemical equation:
2 SO2(g) + O2(g) <=> 2 SO3(g)

Step 2: Write the equilibrium constant expression based on the balanced chemical equation:
Kp = [SO3]^2 / [SO2]^2 * [O2]
Note: The square is applied to the concentrations of SO3 and SO2 because their stoichiometric coefficients are 2.

Step 3: Substitute the given values into the equilibrium constant expression:
Kp = (1.03 g / molar mass of SO3)^2 / [SO2]^2 * (0.170 g / molar mass of O2)
Note: You need to convert the given masses of SO3 and O2 into moles by dividing by their respective molar masses.

Step 4: Rearrange the equation to solve for [SO2]:
[SO2]^2 = (1.03 g / molar mass of SO3)^2 / (Kp * (0.170 g / molar mass of O2))
[SO2] = sqrt((1.03 g / molar mass of SO3)^2 / (Kp * (0.170 g / molar mass of O2)))

Step 5: Calculate the amount of SO2 in grams by multiplying the concentration [SO2] by the molar mass of SO2:
mass of SO2 = [SO2] * molar mass of SO2

Step 6: Evaluate the expression. Plug in the values for molar masses of SO3, O2, and SO2, as well as the given value of Kp and the amounts of SO3 and O2 in grams.

Using this approach, you should be able to determine the grams of SO2 in the mixture.