Question

Imagine that you have a 6.00 L gas tank and a 2.50 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 115 atm, to what pressure should you fill the acetylene tank to ensure that you run out of each gas at the same time?

Answer

There are a lot of assumptions to be made here no matter how right or wrong they may be so let me get them out of the way first.  First off, acetylene is not charged to a tank on it's own, it is usually charged to a tank containing acetone which is used to help dissolve and thus stabilize the acetylene to prevent polymerization.  This would take up additional space in the cylinder and screw with things, and doesn't seem to be in line with the level of difficulty of this problem.  Additionally, acetylene doesn't really follow the ideal gas law to the letter of some other gasses, so it should be using its own non-ideal gas equation, but since you weren't given any of those constants, and again because it doesn't seem to be in line with the level of difficulty of this problem, I would think this is a moot point.  So, first off, let's figure out the equation for the stoichiometry.

2HCCH + 5O₂ ---> 2H₂O + 4CO₂

Now, let's use the ideal gas law to determine how many mols of oxygen you have in your 6.00L cylinder using the parameters outlined in the problem.

PV = nRT
(P = Pressure, V = Volume, n = mols, R = gas constant based on what units you have, T = temp)  Solve for n

n = PV/(RT)

n = (115 atm x 6.00 l)/[(0.082057 L x atm/K x mol) x (293.15 K)
R chosen to fit other units, T chosen to be at STP for 20ºC

n = 690 atm x L / (24.05500955 L x atm / mol)

n = 28.7 mols O₂
Rounded for significant figures

Now that we know how much oxygen there is in the cylinder we can use the balanced equation above to see that for every five mols of O₂ we need 2 mols of acetylene.  So, divide by five then multiply by two to get how many mols of acetylene are needed (note that we are assuming this is a complete combustion and that no atmospheric oxygen is used).

28.7 mols O₂ / (5 mols O₂ / 2 mol acetylene) = 11.48 mol acetylene

Now plug this into the ideal gas equation, this time using the volume of the cylinder that is to contain the acetylene.  You are solving for the pressure and you have all the other variables.