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

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

1 Answer
Oct 12, 2016

Answer:

We need a stoichiometrically balanced equation:

#CH_4(g) + 2O_2(g) rarr CO_2(g) +2H_2O(l)#

Explanation:

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

We started with #4.0*mol# of methane, and thus we require #8.0*mol# 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.