How many liters of oxygen gas are required to react with 2.25 L of methane at 21.0 C and 0.997 atm?

Answer 1

Write the correctly balanced equation for the reaction:

CH₄ + 2O₂ ==> CO₂ + 2H₂O ... balanced equation for combustion of methane

moles CH₄ present:

PV = nRT

n = PV/RT = (0.997 atm)(2.25 L) / (0.0821 Latm/Kmol)(294K)

n = 0.0942 moles

moles O₂ needed = 0.0942 mol CH₄ x 2 mol O₂ / mol CH₄ = 0.1184 moles O₂ needed

Volume O₂ needed:

V = nRT/P = (0.1184)(0.0821)(294K) / (0.997)

V = 4.56 L of O₂ needed