consider the reaction 2so2(g) + o2(g)=2so3(g) taking place at 50.0 c and 1.000 a

Question

consider the reaction 2so2(g) + o2(g)=2so3(g) taking place at 50.0 c and 1.000 atm pressure.

if 300.0 ml of so2(g) react with 150.0 ml of O2(g), what vloume of so3(g) will be produce (at 50.0 c and 1.000 atm pressure)?

Note: The answer is 300.0 ML, but I don't HOW??? I want the steps plz

Explanation / Answer

Pressure (P) = 1.00 atm : Temperature(T) = 50C = 323 K : R = gas constant = 0.0821 L atm/K mol Now for SO2 volume (V) = 300.0mL = 0.3 L According to Ideal gas law PV=nRT n = PV/RT =(1.0*0.3)/(0.0821*323) = 0.0113 mol SO2 Similarly for O2, volume(V) = 150.0mL = 0.15 L PV=nRT n = PV/RT =(1.0*0.15)/(0.0821*323) = 0.00565 mol O2 Now the balanced reaction is 2SO2(g) + O2(g)====>2SO3(g The molar ratio between SO2 and O2 is 2:1 0.0113 mol SO2/2 = 0.000565 mol SO2 0.00565 mol O2/1 = 0.00565 mol O2 So both of the reactant reacts completely to form SO3. Let's take O2 0.00565 mol O2 * (2 mol SO2/1 mol O2) 0.0113 mol SO3 So PV=nRT V= nRT/P = (0.0113*0.0821*323)/1 = 0.3L =300.0mL SO3

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