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Gas Law Stoichiometry,?
2(C2H2) + 5(O2) = 4(CO2) + 2(H2O)
= is an arrow
Imagine that you have a 6.00 L gas tank and a 4.50 L 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 125 atm, 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.
You can use the expression
P1V1 = P2V2
But you have to factor in the stiochiometry of the reaction
P1 = pressure O2 (125 atm)
V1 = volume O2 (6.00L)
P2 = pressure C2H2 (?)
V2 = volume C2H2 (4.50 L)
P2 = P1V1 / V2
2 moles C2H2 needs 5 moles O2 to react
Therefore
P2 = (2/5 x P1V1) / V2
P2 = (2/5 x 125 atm x 6.00 L) / 4.50 L
P2 = 66.7 atm
I have attempted to show why this work below...
PV = nRT
P = pressure
V = volume
n = moles
R = const
T = temp
For the oxygen
n(O2) = P1V1 / RT
So you can write an expression for moles C2H2 based on moles O2
n(C2H2) = 2/5 x n (O2)
= 2/5 x (P1V1 / RT)
Now, sub the expression for moles C2H2 into PV=nRT for C2H2
P2 = nRT / V2
sub in the expression for moles C2H2
P2 = ((2/5P1V1) / RT) x RT / V2
The RT's cancel out (1/RT x RT = 1) and you are left with
P2 = (2/5 x P1V1) / V2
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