Question

Under standard conditions, what volume of carbon dioxide gas results from combustion of a `19.8*g` mass of methane?

Answer 1

Well one mole of gas occupies `22.4*L` at STP....

Explanation:

You have gots the stoichiometric reaction.....

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

And with respect to methane there is a `(19.8*g)/(16.01*g*mol^-1)=1.24*mol`....

And upon combustion we necessarily get a `1.24*mol` quantity of carbon dioxide.

Now by the given stoichiometry, there must necessarily be a `2.48*mol` quantity of dioxygen....

And this occupies a volume of........... `2.48*molxx22.4*mol*L^-1~=55*L`