Question

HELLLPPPP!!!
How many liters of methane gas (CH4) need to be combusted to produce 8.5 liters of water vapor, if all measurements are taken at the same temperature and pressure? Show all of the work used to solve this problem.

CH4 (g) + 2 O2 (g) yields CO2 (g) + 2 H2O (g)

Answer 1

well
are you familiar with STP?

Answer 2

We can figure out this question by using the ideal gas law, PV=nRT.

P = pressure
V = volume
n = moles
R = gas constant
T = temperature

We have two situations here: your initial conditions, i (before the reaction starts), and your final conditions, f (after the reaction finishes). Therefore we can say:

\[\frac{ P_{i}V_{i} }{ n_{i}T_{i} }=\frac{ P_{f}V_{f} }{ n_{f}T_{f} }\]
All I've done here is rearranged PV=nRT to R = PV/nT, and set the initial and final equations equal to each other (since R is a constant and so is always the same). The question tells us that pressure and temperature stay the same the entire time, so we can take those out of the equation to get:
\[\frac{ V_{i} }{ n_{i} }=\frac{ V_{f} }{ n_{f} }\]
This, by the way, is called Avogadro's Law. We can do one more thing with this equation though - we can say the left side of our equation represents CH4, and the right side represents H2O. This means n(i) is the moles of CH4, and n(f) is the moles of H2O. If you look at your chemical reaction, you'll see that there are 2 H2O produced for every CH4 consumed. This means that n(f) = 2n(i). Are equation becomes:

\[\frac{ V_{i} }{ n_{i} }=\frac{ V_{f} }{ 2n_{i} }\]\[ V_{i} =\frac{ V_{f} }{ 2 }\]
This equation makes sense - since all the other factors (i.e. pressure and temperature) remain the same, twice as many moles occupy twice the volume. To find your starting volume of CH4, just divide the volume of H2O by 2 to get 4.25 L!

If you have any questions please ask