Question

If we completely combust a `2.40xx10^-3*g` mass of methane, what is the mass of water evolved?

Answer 1

Approx. `5*mg` . How do we know.........?

Explanation:

You should be able to represent these combustion reactions by means of an equation........

`"For methane........."`

`CH_4(g) + O_2(g) rarr CO_2(g) + 2H_2O(l)+Delta_1`

`"For ethane........."`

`C_2H_6(g) + 7/2O_2(g) rarr 2CO_2(g) + 3H_2O(l)+Delta_2`

Now not only are atoms, and MASS stoichiometrically balanced, i.e. `"garbage in garbage out"`, but enthalpy/energy are also stoichiometrically balanced.

We combust a molar quantity of `(2.40xx10^-3*g)/(16.04*g*mol^-1)=1.50xx10^-4*mol` with respect to methane........

And given the stoichiometry of the reaction........we get `1.50xx10^-4xx2xx18.01*g*mol^-1=??g` with respect to water....

Given combustion of EQUIMOLAR methane and ethane, would you predict that `Delta_2>Delta_1`? Why or why not?