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 105 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.

Express your answer with the appropriate units.

Answer 1

2C2H2(g) + 5O2---------> 4CO2 + 2H2O

Two moles of acetylene reacts with five moles of oxygen.

Oxygen Data:

V = 6.00 L

P = 105 atm

T = 298 K

R = 0.08205 LatmK^-1mol^-1

using ideal gas equation:

PV=nRT (Ideal gas equation)

n = PV/RT

n = 105 atm x 6.0 L / 0.08205 LatmK^-1mol^-1 x 298 K

n = 25.76 moles

Acetylene Data:

25.76 moles correspond to 5 moles of oxygen, so the required amount of acetylene for complete combustion is : (2x25.76)/5 =10.3 moles

V = 2.50 L

P = ?

T = 298 K

R = 0.08205 LatmK^-1mol^-1

using ideal gas equation:

PV=nRT

P = nRT / V

P = 10.3 x 0.08205 LatmK^-1mol^-1 x 298 K / 2.50 L

P = 100.73 atm

So the required pressure of acetylene tank is 100.73 atm