Asked by James
Sulfur trioxide, SO3, is produced in enormous quantities each year for use in the synthesis of sulfuric acid.
S(s) + O2(g) -> SO2(g)
2SO2(g) + O2(g) -> 2SO3(g)
What volume of O2 (g) at 350 degrees Celsius and a pressure of 5.25 atm is needed to completely convert 5.00g of sulfur to sulfur trioxide?
S(s) + O2(g) -> SO2(g)
2SO2(g) + O2(g) -> 2SO3(g)
What volume of O2 (g) at 350 degrees Celsius and a pressure of 5.25 atm is needed to completely convert 5.00g of sulfur to sulfur trioxide?
Answers
Answered by
DrBob222
The long way, but perhaps easier to understand, is to convert 5 g S to moles, then to mole O2 needed and moles SO2 produced in the first equation. Take the mole SO2 produced in equation 1 and use that to determine moles oxygen from equation 2. Add the two to find total moles O2. Then use PV = nRT to find volume O2. Check my thinking.
Answered by
MikeKhajiit
Okay, I know this is 10 years old. I just found it because I had this exact same problem on a homework assignment. Here's what I got after a bit of work:
V = Volume of O2
350.°C -> 623 K
5.00 g S / 32.06 g S = 0.156 mol S
0.156 / 2 = 0.078 mol O2
0.156 + 0.078 = 0.234 mol O2
V = (0.234 * 0.08206 * 623) / 5.25 = 2.28 L O2
I'm pretty sure this is right. I hope it helps whoever comes next!
V = Volume of O2
350.°C -> 623 K
5.00 g S / 32.06 g S = 0.156 mol S
0.156 / 2 = 0.078 mol O2
0.156 + 0.078 = 0.234 mol O2
V = (0.234 * 0.08206 * 623) / 5.25 = 2.28 L O2
I'm pretty sure this is right. I hope it helps whoever comes next!
Answered by
MikeKhajiit
I forgot to include the equations I used.
I used this one:
PV = nRT
But we were solving for volume, so I changed it to this:
V = nRT / P
I used this one:
PV = nRT
But we were solving for volume, so I changed it to this:
V = nRT / P
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