# What mass of water results from combustion of an 8.0*mol quantity of "methane gas"?

Nov 17, 2016

Under $300 \cdot g$ water.

#### Explanation:

We need a stoichiometrically balanced equation to represent the combustion:

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

The stoichiometry dictates that for each mole of methane, 2 moles of water are evolved.

So if $8.0 \cdot m o l$ methane gas were combusted, $16 \cdot m o l$ of water were evolved: 16*molxx18.02*g*mol^-1=???g