The first step is to write a stoichiometrically balanced equation that represents the COMPLETE combustion of the hydrocarbon:
#CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(l)+Delta#
And please note that the heat output, #Delta#, could also be accounted for stoichiometrically. And then we work out the equivalent quantities of methane, which is the limiting reagent.
#"Moles of methane"=(96*g)/(16.01*g*mol^-1)=6*mol.#
Given the stoichiometric equation, CLEARLY, we get #6*mol# of carbon dioxide gas upon combustion, which represents a mass of.....
Are you happy with this?