Assume that the partial pressure of sulfur dioxide, pso2, is equal to the partial pressure of dihydrogen sulfide, ph2s, and therefore pso2=ph2s. if the vapor pressure of water is 22 torr , calculate the equilibrium partial pressure of so2 (pso2) in the system at 298 k. express the pressure in atmospheres to two significant figures.

First you need to know the value of Kp for this reaction. The reaction is: SO2 (g) + 2 H2S (g)<=> 3S(s) + 2 H2O(g): Kp = PH2O^2 / PSO2(PH2O)^2 but since the problem is lacking in details.
Otherwise, let x = PSO2 = PH2S. Then:
Kp = (22/760 atm)^2 / x^3 
x = cube root (0.289)/Kp = PSO2
But if the kp is given, let us assume it is 8.0*10^15The solution would be:convert the given pressure of water to atm, which is: 0.029
Kp=P(H2O)^2/P(SO2)P(H2S)^2
8.0*10^15 = (0.029)^2/x^3
(8.0*10^15).(x^3)=8.41*10^-4
take the cube root of both sides(2.0*10^5).x = 9.439130*10^-2x = P (SO2) = 4.6*10^-7

Ibelieve y y even though i've only taken chemistry for one year

We can't until we know what we are working with or on