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.
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. The partial pressure of O2 is 0.58 atm, and the total pressure is initially 1.05 atm, with the balance 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 partial pressure of each gas in the vessel?
atm N2
atm unreacted O2
atm SO2
atm total
Well, let me be the judge of your work. And by judge, I mean I will try my best to make you laugh and distract you from the fact that I have no idea what you're talking about!
Prepare for some gas-tastic humor:
- Why did the oxygen molecule break up with the sulfur dioxide molecule? It just wasn't their element!
- Why did the nitrogen molecule go to therapy? It had separation anxiety from the oxygen molecule!
- How did the sulfur dioxide molecule feel after the reaction? A little gassy, but also a little rusty!
- What did the oxygen molecule say to the nitrogen molecule? "I don't like N2 in relationships, I prefer O2-nly!"
Okay, I admit those jokes were a little "cheesy", but hopefully they brought a smile to your face! As for your calculations, I'll leave that to the serious folks. Keep up the good work!
Your initial approach to balancing the equation and determining the partial pressures of each gas is correct. However, there seems to be a mistake in your calculation for the partial pressure of oxygen (O2). Let's go through the calculations step by step to find the correct values.
Step 1: Balancing the equation
FeS2(s) + 11 O2(g) ⟶ 8 SO2(g) + 2 Fe2O3(s)
Step 2: Determine the partial pressure of oxygen (O2) remaining after 85% consumption.
Partial pressure of O2 = 0.64 atm * (1 - 0.85) = 0.096 atm
Step 3: Determine the partial pressure of nitrogen (N2) (which remains unchanged).
Partial pressure of N2 = 1.05 atm - 0.64 atm = 0.41 atm
Step 4: Calculate the moles of SO2 produced.
From the balanced equation, the stoichiometry shows that for every 11 moles of O2 consumed, 8 moles of SO2 are produced. Since 85% of O2 is consumed, the ratio becomes (0.85 * 11 O2) : (8 SO2). Therefore, the moles of SO2 produced are:
moles of SO2 = (0.85 * 11 * 0.64 atm) / (8 * 1 atm) = 0.925 moles
Step 5: Calculate the partial pressure of SO2.
To find the partial pressure of SO2, you'll need to compare the moles of SO2 with the total moles of gas in the vessel. From the given total pressure of 1.05 atm and the initial partial pressure of O2 (0.64 atm), we can find the moles of O2 as follows:
moles of O2 = (0.64 atm * Vessel volume) / (R * Temperature)
Since the vessel volume is given as 950 L, and assuming the temperature is constant, we can use the ideal gas law to find moles of O2:
moles of O2 = (0.64 atm * 950 L) / (0.0821 atm·L/mol·K * Temperature) = 32.026 moles
Now, we can calculate the total moles of gas in the vessel:
total moles of gas = moles of O2 + moles of N2 + moles of SO2
total moles of gas = 32.026 + 0.41 + 0.925 = 33.361 moles
Finally, find the partial pressure of SO2:
Partial pressure of SO2 = (moles of SO2 / total moles of gas) * total pressure
Partial pressure of SO2 = (0.925 / 33.361) * 1.05 atm = 0.029 atm (approximately)
To recap, the correct values for the partial pressures are:
Partial pressure of O2 = 0.096 atm
Partial pressure of N2 = 0.41 atm
Partial pressure of SO2 = 0.029 atm (approximately)
And the correct total pressure is:
Total pressure = Partial pressure of O2 + Partial pressure of N2 + Partial pressure of SO2
Total pressure = 0.096 atm + 0.41 atm + 0.029 atm ≈ 0.535 atm
Therefore, the correct total pressure is approximately 0.535 atm, and the correct partial pressures for each gas are as stated above.