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

Jul 7, 2017

Approx. $5 \cdot m g$ . How do we know.........?

#### Explanation:

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

$\text{For methane.........}$

$C {H}_{4} \left(g\right) + {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right) + {\Delta}_{1}$

$\text{For ethane.........}$

${C}_{2} {H}_{6} \left(g\right) + \frac{7}{2} {O}_{2} \left(g\right) \rightarrow 2 C {O}_{2} \left(g\right) + 3 {H}_{2} O \left(l\right) + {\Delta}_{2}$

Now not only are atoms, and MASS stoichiometrically balanced, i.e. $\text{garbage in garbage out}$, but enthalpy/energy are also stoichiometrically balanced.

We combust a molar quantity of $\frac{2.40 \times {10}^{-} 3 \cdot g}{16.04 \cdot g \cdot m o {l}^{-} 1} = 1.50 \times {10}^{-} 4 \cdot m o l$ 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?