Question #a560b

1 Answer
anor277
May 28, 2017

We gets over #250*g# #CO_2(g)#.

Explanation:

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.....

#6*molxx44.01*g*mol^-1=??g#

Are you happy with this?

Related questions
See all questions in Stoichiometry
Impact of this question
1316 views around the world
You can reuse this answer
Creative Commons License