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

1 Answer
Jul 7, 2017

Answer:

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?