given the chemical equation, 2C2H6(s)+7O2(g)-->4CO2(g)+6H2O(g), determine the volume in liters of CO2 produced from 52.53 L of O2 at STP

Answer 1

2C₂H₆(l) + 7O₂(g) ==> 4CO₂(g) + 6H₂O(l) .. balanced equation

At STP, we can use liters in place of moles in interpreting the balanced equation. Thus, we can read this as

"2 liters of C₂H₆ reacts with 7 liters of O₂ to produce 4 liters of CO₂".

Now, using this stoichiometry, we can solve the problem:

52.53 L O₂ x 4 L CO₂ / 7 L O₂ = 30.02 L CO₂

If you don't trust this approach, do it the long way using the ideal gas law:

PV = nRT

P = 1 atm (STP)

V = 52.53 L

n = moles of O₂ = ?

R = gas constant = 0.0821 Latm/Kmol

T = 273K (STP)

n = PV/RT = (1)(52.53) / (0.0821)(273)

n = 2.34 mols O₂

Now use the stoichiometry of the balanced equation to find moles of CO2 produced:

2.34 mol O₂ x 4 mols CO₂ / 7 mols O₂ = 1.34 mols CO₂ produced

Convert this to volume of CO₂:

V = nRT / P = (1.34 mol)(0.0821)(273) / 1

V = 30.03 L CO₂