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) Imagine that you have a 7.00 L gas tank and a 2.00 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 155 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.

Answer 1

Answer – We are given , Pressure of O₂ = 155 atm, volume = 7.00 L ,

Volume of acetylene = 2.00 L, Pressure of acetylene = ?

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

From the balanced reaction 5 moles of O₂ needed 2 moles of C₂H₂(g)

Now we need to calculate moles of O₂

We know

PV = nRT

So, n = PV/ RT

           = 155 atm * 7.00 L / 0.0821 L.atm. mol^-1.K^-1 *298 K

           = 44.34 moles of O₂

From the above reaction –

5 moles of O₂ = 2 moles of C₂H₂(g)

44.34 moles of O₂ = ?

= 17.76 moles of C₂H₂(g)

Now pressure of C₂H₂(g)-

PV = nRT

P = nRT/V

   = 17.76 mole x 0.0821 L.atm. mol^-1.K^-1 *298 K / 2.00 L

= 217 atm

217 atm pressure should you fill the acetylene tank to ensure that you run out of each gas at the same time.