# Assuming all volume measurements are made at the same temperature and pressure, what volume of hydrogen gas is needed to react completely with 4.55 L of oxygen gas to produce water vapor?

##### 1 Answer

#### Explanation:

The important thing to keep in mind when dealing with gases that are **under the same conditions for pressure and temperature** is that their mole ratio is **equivalent** to their *volume ratio*.

You can prove this by using the ideal gas law equation

#P * V_1 = n_1 * RT -># the first gas at pressure#P# and temperature#T#

#P * V_2 = n_2 * RT -># the second gas at pressure#P# and temperature#T#

Divide these two equations to get

#(color(red)(cancel(color(black)(P))) * V_1)/(color(red)(cancel(color(black)(P))) * V_2) = (n_1 * color(red)(cancel(color(black)(RT))))/(n_2 * color(red)(cancel(color(black)(RT))))#

Therefore, you can say that

#n_1/n_2 = V_1/V_2 -># the mole ratio is equal to the volume ratio

The balanced chemical equation for your reaction looks like this

#color(red)(2)"H"_text(2(g]) + "O"_text(2(g]) -> 2"H"_2"O"_text((g])#

Notice that you have a **moles** of hydrogen gas **for every** mole of oxygen gas that takes part in the reaction.

Now, since all three gases are under the same conditions for *pressure* and *temperature*, this mole ratio will be equal to a **volume ratio**

#n_(H_2)/n_(O_2) = V_(H_2)/V_(O_2) = color(red)(2)/1#

This means that your sample of oxygen would need

#4.55 color(red)(cancel(color(black)("L O"_2))) * (color(red)(2)" L H"_2)/(1color(red)(cancel(color(black)("L O"_2)))) = color(green)("9.10 L H"_2)#

The answer is rounded to three sig figs.