At a given temperature, an 8*mol8mol quantity of methane that occupies a volume of 200*L200L is combusted with stoichiometric oxygen gas. Can you address...?

"(i) the molar quantity of dioxygen gas required for equivalence;"(i) the molar quantity of dioxygen gas required for equivalence;
"(ii) the volume of dioxygen gas required for equivalence;"(ii) the volume of dioxygen gas required for equivalence;

1 Answer
Oct 12, 2016

We need a stoichiometrically balanced equation:

CH_4(g) + 2O_2(g) rarr CO_2(g) +2H_2O(l)CH4(g)+2O2(g)CO2(g)+2H2O(l)

Explanation:

The balanced chemical equation tells us that "1 mole/1 equiv"1 mole/1 equiv of methane gas requires "2 moles/2 equiv"2 moles/2 equiv of dioxygen gas for complete combustion.

We started with 4.0*mol4.0mol of methane, and thus we require 8.0*mol8.0mol of dioxygen gas to combust this completely. This represents a mass of 4.0*cancel(mol)xx16.01*g*cancel(mol^-1)=64.0*g methane, and how many grams of dioxygen gas?

Since the volume is specified, there thus must be TWICE the volume of dioxygen compared to methane for stoichiometric equivalence.