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 7.00 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 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

Given reaction,

2C2H2(g) + 5O2(g)→4CO2(g) + 2H2O(g)

This means 2 moles of C2H2 require 5 moles of O2 for complete combustion

For oxygen,

Volume = V1= 7.00 L

Pressure = P1= 155 atm.

n1 = 5 mole

For acetylene,

V2 = 2.50L

P2 = ?

n2 = 2

As per ideal gas law equation,

PV = n R T

RT = P V /n

Where,

R = gas constant

T = consider both gas filling was done at same temperature = constant.

For given reaction,

(P1 × V1)/n1 = (P2 × V2)/n2

Putting above values,

(155atm × 7.00L)/5mol = (P2 × 2.50L)/2mol

P2 = (155atm × 7.00L × 2 mol)/ (5mol × 2.50L)

P2 = [(155 × 7.00 × 2) /(5 × 2.50)]atm

P2 = 173.6atm.

Pressure of acetylene tank to be filled to ensure that you run out of each gas at the same time = 173.6atm.