What volume of carbon dioxide gas will be produced when 5.45 L of methane and 7.35 L of oxygen gas are allowed to react at 30.0 Celsius and 1.02 bar?

Question

Balanced chemical equation:

CH₄ (g) + 2 O₂ (g) --> CO₂ (g) + 2 H₂O (g)

Answer 1

CH₄ (g) + 2 O₂ (g) --> CO₂ (g) + 2 H₂O (g) ... balanced equation

Because the temperature and pressure are constant at 30ºC and 1.02 bar respectively, the stoichiometry of the balanced equation can be used in terms of moles, or volume (liters). Thus, according to the balanced equation, you need twice as much O₂ as CH₄. Since you have only 7.35 L O₂, that isn't twice the volume of CH₄, so O₂ becomes limiting.

7.35 L O₂ x 1 L CO₂ / 2 L O₂ = 3.675 L CO₂ = 3.68 L CO₂ (3 sig. figs.)

If you wanted to do it the long way, you can use the ideal gas law: PV = nRT

P = 1.02 bar x 0.987 atm / bar = 1.00 atm

R = 0,0821 Latm/Kmol

T = 30 + 273 = 303K

n = moles = PV/RT

moles CH₄ = (1.00 atm)(5.45 L) / (0.0821 Latm/Kmol)(303K) = 0.219 moles

moles O₂ = (1.00 atm)(7.35 L) / (0.0821 Latm/Kmol)(303K) = 0.295 moles

Limiting reactant = O₂ since you need twice as many mols O₂ as CH₄ (see balanced equation)

The moles of the limiting reactant (O₂) will determine moles of CO₂ than can be formed...

0.295 mols O₂ x 1 mol CO₂ / 2 mol O₂ = 0.1475 moles CO₂

Volume CO₂ = nRT/P = (0.1475 mol)(0.0821)(303) / 1.00 atm = 3.68 L CO₂