# 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?

Dec 26, 2015

$\text{9.10 L}$

#### 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

$P \cdot {V}_{1} = {n}_{1} \cdot R T \to$ the first gas at pressure $P$ and temperature $T$

$P \cdot {V}_{2} = {n}_{2} \cdot R T \to$ the second gas at pressure $P$ and temperature $T$

Divide these two equations to get

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{P}}} \cdot {V}_{1}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{P}}} \cdot {V}_{2}} = \frac{{n}_{1} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{R T}}}}{{n}_{2} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{R T}}}}$

Therefore, you can say that

${n}_{1} / {n}_{2} = {V}_{1} / {V}_{2} \to$ the mole ratio is equal to the volume ratio

The balanced chemical equation for your reaction looks like this

$\textcolor{red}{2} {\text{H"_text(2(g]) + "O"_text(2(g]) -> 2"H"_2"O}}_{\textrm{\left(g\right]}}$

Notice that you have a $\textcolor{red}{2} : 1$ mole ratio between hydrogen gas and oxygen gas. This mole ratio tells you that you need $\textcolor{red}{2}$ 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

${n}_{{H}_{2}} / {n}_{{O}_{2}} = {V}_{{H}_{2}} / {V}_{{O}_{2}} = \frac{\textcolor{red}{2}}{1}$

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.