Question

Question #a560b

Answer 1

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?