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
You need to work out the concns.
(SO3) = (1.11g/molar mass SO3)/2L = approx 0.007M but you need to do it more accurately as well as calculations that follow. They are estimates also.
(O2) = 1.06/32/2L = approx 0.0015M
Kc = (SO3)^2/(SO2)^2(O2)
Substitute Kc and concns SO3 and O2 and solve for (SO2)
After (SO2) in M, then
mol = M x L
Then mols = grams/molar mass. Solve for grams.
To find the number of grams of SO2 present in the vessel, we need to use the given information about the equilibrium reaction and the amounts of SO3 and O2 present.
First, let's convert the masses of SO3 and O2 to moles using their molar masses. The molar mass of SO3 is 80.07 g/mol, so:
Number of moles of SO3 = Mass of SO3 / Molar mass of SO3
= 1.11 g / 80.07 g/mol
= 0.013857 mol
The molar mass of O2 is 32.00 g/mol, so:
Number of moles of O2 = Mass of O2 / Molar mass of O2
= 0.106 g / 32.00 g/mol
= 0.003313 mol
Since the balanced equation shows that the ratio of SO2 to O2 is 2:1, we know that the number of moles of SO2 is twice the number of moles of O2.
Number of moles of SO2 = 2 * Number of moles of O2
= 2 * 0.003313 mol
= 0.006626 mol
Finally, we can convert the number of moles of SO2 to grams using the molar mass of SO2, which is 64.06 g/mol:
Mass of SO2 = Number of moles of SO2 * Molar mass of SO2
= 0.006626 mol * 64.06 g/mol
= 0.4244 g
Therefore, there are approximately 0.4244 grams of SO2 present in the vessel.