Question

Imagine that you have a 7.00 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 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

Step 1: Write the balanced chemical equation

2 C₂H₂ + 5 O₂ ------>  4 CO₂ + 2 H₂O

Short method 1: Step 2: Calculation

2 C₂H₂ + 5 O₂ ------>  4 CO₂ + 2 H₂O

From the equation we can see

2 moles of acetylene are needed for 5 moles of oxygen

which means we need 2.5 times as much oxygen for  acetylene

P₂ = mole ratio of acetylene × (P₁V₁) / V₂

=> P₂ = 2/5 × (155 atm × 7 L) / 3 L = 144.67 atm

Method 2:

step 1: calculate the moles of oxygen

Assume both are operating at 25 °C  which means T = (273 + 25) = 298 K

pressure = 155 atm and volume = 7 L

By using ideal gas equation

PV = nRT

=> n = PV/RT = (155 atm × 7 L ) / (0.08206 L.atm /mol.K × 298 K ) = 44.37 mol

Step 2: Calculate the moles of acetylene

2 C₂H₂ + 5 O₂ ------>  4 CO₂ + 2 H₂O

From the equation we can see

5 moles of oxygen  are needed for 2 moles of acetylene

so for 44.37 mol of oxygen we need = ( 2 mol / 5 mol )  × 44.37 mol = 17.748 mol

Step 3: calculate the pressure of acetylene

moles(n) = 17.748 mol

Volume = 3 L and Temperature(T) = 298 K

By using ideal gas equation

PV = nRT

=> P = nRT / V = ( 17.748 mol  × 0.08206 L.atm /mol.K × 298 K ) / 3 L = 144.67 atm