Gaseous Chemical Reactions and Volumes

[danielakkerma]

(Hi everyone! Apologising for the trivial(and likely boring) question in advance. Sadly, it has me boggled for some reason).

Homework Statement

In a certain temperature and under a pressure, 500 mL of H2, and 100 mL of O2 are poured into a container. A Chemical reaction occurs as follows:
$$2H_2+O_2 \rightarrow 2H_2O (g)$$
Find the volume of the water vapour released in the reaction, with the same pressure and temperature.

Homework Equations

(My physical inclination): Conservation of Mass...

The Attempt at a Solution

Since no mass, of any element, is lost during this process, the right-hand and left-hand sides of the equation must equal, when considering their respective mass input.
So:
$$2\cdot x\cdot m^{'}_{H_2}+y\cdot m^{'}_{O_2} = 2\cdot z\cdot m^{'}_{H_2O}$$
(m' denotes the respective molar mass. x, y, z, mole counts, derived below).
Since I know the Physical conditions to be constant during process, I take arbitrary Pressure, and Temperature (P, T), to obtain:
$$x_i = \frac{pV_i}{RT}$$
Therefore:
$$2 \frac{pV_1}{RT}m^{'}_{H_2}+\frac{pV_2}{RT}m^{'}_{O_2}=2\frac{pV_3}{RT}m^{'}_{H_2O}$$
As expected, I lose my P/RT, to find:
$$2 m^{'}_{1}\cdot V_1+m^{'}_{2}\cdot V_2 = 2m^{'}_{3}\cdot V_3$$
Therefore, solving for V_3:
$$V_3 = \frac{2 m^{'}_{1}\cdot V_1+m^{'}_{2}\cdot V_2}{2m^{'}_{3}} \approx 144.44 mL \\ m^{'}_{1} \Rightarrow H_2 = 2\frac{g}{mol} \\ m^{'}_{2} \Rightarrow O_2 = 32\frac{g}{mol} \\ m^{'}_{3} \Rightarrow H_2O = 18\frac{g}{mol}$$
Yet, I fear, this is not the correct answer!
What have I done wrong?
Thankful, as always, for your attention,
Daniel

 

[danielakkerma]

Sorry guys and gals for bumping this, but I've ascertained the correct answer to be 200 mL and I still can't figure out how to get it!
Any help would be greatly appreciated,
Beholden,
Daniel

 

[danielakkerma]

Okay folks, solution at hand!
My mistake was rooted in the fact that I didn't consider who the limiting reagent was, in this question.
By calculating with the given volumes I find that $$\frac{n_{H_2}}{n_{O_2}} = 5 > 2$$(n -> number of moles), meaning that O2 will form my constraint. It will have run out before the sequestration of all the Hydrogen in the system.
Therefore, the amount of Water vapour produced will depend solely on the amount of Oxygen invested in the reaction.
Thus: $$n_{O_2} = \frac{PV_{O_2}}{RT} \\ 1 n_{O_2} \rightarrow 2 n_{H_2O} \\ \frac{2PV_{O_2}}{RT}m^{'}_{H_2O} = \rho_{H_2O}V_{H_2O} \text{ (Conservation of mass)} \\ \rho = \frac{m^{'}P}{RT} \\ \frac{2PV_{O_2}}{RT}m^{'}_{H_2O} = \frac{m^{'}_{H_2O}P}{RT}V_{H_2O} \\$$
Pm'(H20)/RT is lost, to finally yield:
$$V_{H_2O} = 2V_{O_2} = 200 mL$$ which is indeed the right answer!
Thanks for your attention everyone,
And I hope this helps someone else, with similar difficulties.

 

[Borek]

Limiting reagent first, but then you did a lot of completely unnecessary work - Avogadro's law tells us ratio of volumes (measured at the same P,T) is identical to the ratio of numbers of moles. This makes the exercise trivial.

 

[danielakkerma]

Thanks Borek... that'll really come in handy.
(The Mathematical sirens in me decided to lure me into a pointless exercise in variables and &c). :D.
Of course, in hind sight, applying Avogadro is the only rational way to go, as you say.
Will keep it simpler, from now on.
Thank you!