# Question a560b

May 28, 2017

We gets over $250 \cdot g$ $C {O}_{2} \left(g\right)$.

#### Explanation:

The first step is to write a stoichiometrically balanced equation that represents the COMPLETE combustion of the hydrocarbon:

$C {H}_{4} \left(g\right) + 2 {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right) + \Delta$

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.

$\text{Moles of methane} = \frac{96 \cdot g}{16.01 \cdot g \cdot m o {l}^{-} 1} = 6 \cdot m o l .$

Given the stoichiometric equation, CLEARLY, we get $6 \cdot m o l$ of carbon dioxide gas upon combustion, which represents a mass of.....

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

Are you happy with this?