# Chemistry

Imagine that you have a 7.00L gas tank and a 3.50L gas tank. You need to fill one tank with oxygen and the other with acetylene to use in conjunction with your welding torch. If you fill the larger tank with oxygen to a pressure of 135atm , to what pressure should you fill the acetylene tank to ensure that you run out of each gas at the same time? Assume ideal behavior for all gases.

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3. 👁 5,766
1. 2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g) formula for above

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2. PV = nRT for the 7 L tank for O2 @ 135 atm and solve for n.
Use the equation to convert mols O2 to mols C2H2.
Use PV = nRT and n mols C2H2 and solve for P in the smaller tank.

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3. I don't know the answer, I don't know how to do it, and your answer didn't get me any closer to solving it.

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4. I didn't work it for you but I gave you the three steps to solve it. It isn't easy to help if you don't tell me what you don't understand.
Three steps. Copied from above.

1. PV = nRT for the 7 L tank for O2 @ 135 atm and solve for n.
2. Use the equation to convert mols O2 to mols C2H2.
3. Use PV = nRT and n mols C2H2 and solve for P in the smaller tank.

I'll start.
Step 1 from above.
PV = nRT
P = 135 atm
V = 7 L
n = ?
R = 0.08206
T = isn't given so use any T but use the same T throughout. An easy choice would be 300 K.
135*7 = n*0.08206*300.
Solve for n. You do that first.
What is n?

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5. 38.37

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6. Good.
Now use the coefficients in the balanced equation to convert mols O2 to mols C2H2. That is
38.38 mols O2 x (2 mols C2H2/5 mols O2) = 38.38 mols x 2/5 = about 15 but you can clean up the numbers.
Now use the smaller tank and PV = nRT and solve for P.
P = ?
V = 3.50 L
R = same as above.
T = 300 if that's what you used before.

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7. 107.95 atm

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8. I think you're allowed only 3 significant figures so I would round that to 108,
Look through your problem and make sure no temperature was given.
I must leave the house. Back in about an hour.

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