Question

Acetylene torches are used for welding. These torches use a mixture of acetylene gas, \rm C_2H_2, and oxygen gas, \rm O_2 to produce the following combustion reaction:

\rm 2C_2H_2{\it (g)} + 5O_2{\it (g)} \rightarrow 4CO_2{\it (g)} + 2H_2O{\it (g)}
Imagine that you have a 5.50 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 135 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

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

According to the balanced equation ,

2 moles of C₂H₂ reacts with 5 moles of O₂

From ideal gas laws PV = nRT

Where P - Pressure , V- volume , R - gas constant , T - absolute temperature , n - no.of moles

As T & R constant , PV α n

--> PV / n = P'V' / n'

Where

P = Pressure of Oxygen = 135 atm

V = volume of Oxygen = 5.50 L

n = no.of moles of Oxygen = 5 moles

P' = Pressure of acetylene = ?

V' = volume of acetylene = 4.50 L

n' = no.of moles of acetylene = 2 moles

Plug the values we get P' = PVn' / nV'

                                       = ( 135 x 5.50 x 2 ) / ( 5 x 4.5)

                                       = 66 atm

Answer 2

Step1 Moles of O2 = PV/RT = (5.5x135)/RxT

Step2 Moles of C2H2 = P x4.5/RT = (5/2) x(5.5x135/RT)

Step3 P= (5/2) x(5.5x135)/4.5 = 412.5 atm

Answer 3

136 atm