Question

Acetylene torches are used for welding. These torches use a mixture of acetylene gas, C2H2, and oxygen gas, O2 to produce the following combustion reaction: 2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)

Part A Imagine that you have a 6.50 L gas tank and a 3.00 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(g)→4CO2(g)+2H2O(g)

2 moles of acetylene reacts with 5 moles of oxygen gas

Oxygen Tank

V = 6.5 L

P = 105 atm

T = 298 K (Assumed)

R = 0.008205 LatmK^-1mol^-1

PV = nRT (Ideal gas equation)

n = (PV)/ (RT) = (105 atm x 6.5 L) / (0.008205 LatmK^-1mol^-1 x 298 K)

n = 27.91 moles

27.91 moles correspond to 5 moles of oxygen. For the complete combustion of acetylene gas, the number of moles acetylent gas required is : (2 x 27.91) / 5 = 11.16 moles

Acetylene Tank

V = 3.0 L

P = ?

T = 298 K (Assumed)

R = 0.008205 LatmK^-1mol^-1

PV = nRT (Ideal gas equation)

P = (nRT)/V = (11.16 mol x 0.008205 LatmK^-1mol^-1 x 298 K)/ 3.0L

P = 90.95

So the required pressure of acetylene gas is 90.95 atm