Question

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

Answer 1

Under `300*g` water.

Explanation:

We need a stoichiometrically balanced equation to represent the combustion:

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

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

So if `8.0*mol` methane gas were combusted, `16*mol` of water were evolved: `16*molxx18.02*g*mol^-1=???g`