posted by Tori on .
Sulfur dioxide is used to make sulfuric acid. one method of producing it is by roasting mineral sulfides, for example, FeS2(s) + O2 (g) ---> SO2(g) + Fe2O3(s) (unbalanced).
A production error leads to the sulfide being placed in a 950-L vessel with insufficient oxygen. Initially, the partial pressure of O2 is 0.64atm, and the total pressure is 1.05 atm, with the balance due to N2. The reaction is run until 85% of the O2 is consumed, and the vessel is then cooled to its initial temperature. What is the total pressure and the partial pressure of each gas in the vessel?
Can someone check my work??
Initial pressure of O2 is 61%, N2 is 39%.
After balancing my equation I get
4 FeS2(s) + 11 O2 (g) ---> 8 SO2(g) + 2 Fe2O3(s)
If 85% of O2 is consumed, 15% remains as a gas. I believe that 15% of the original .64 pressure = .096atm.
The N2 was not changed, so NO2 = .41 atm still.
The added gas is SO2. Is the pressure the following: 85% .64atm (8 SO2/11 SO2) = .396 atm?
My ending pressures are:
O2 = .10 atm
N2 = .41 atm
SO2 = .40 atm
Total pressure = .91 atm
I agree with the 0.396 atm = PSO2 formed.
I agree with the 0.096 atm = PO2 remaining.
I agree with PN2 = 0.41 atm.
I am confused about two things.
1. Does the N2 form NO2 as you have noted? If so then is part of the 85% O2 consumed used to form NO2? If so that means neither you or I have taken that into account.
2. I think I see now that you have rounded the 0.096 to 0.1; however, you are allowed two s.f. and I would keep the 0.096. Then Ptotal = 0.096 + .40 + 0.41 = 0.906 and that rounds to 0.91 atm as you have it.