Question #744bf

Jul 15, 2015

Both reactants will be completely consumed.

Explanation:

Take a look at the balanced chemical equation for this synthesis reaction

$\textcolor{red}{2} {H}_{2 \left(g\right)} + {O}_{2 \left(g\right)} \to 2 {H}_{2} {O}_{\left(l\right)}$

Notice that you have a $\textcolor{red}{2} : 1$ mole ratio between hydrogen gas and oxygen gas. This means that, regardless of how many moles of hydrogen gas react, you'll need half as many moles of oxygen in order for the reaction to take place.

If you take into account the fact that both gases are under the same conditions for temperature and pressure, the aforementioned mole ratio will become volume ratio.

That is, the two gases must react in a $\textcolor{red}{2} : 1$ volume ratio in order for the reaction to take place.

Here's why that is so. Assuming that you're at a pressure of 1 atm and at a temperature of ${20}^{\circ} \text{C}$, you can use the ideal gas law equation to write

$P {V}_{{H}_{2}} = {n}_{{H}_{2}} R T$

and

$P {V}_{{O}_{2}} = {n}_{{O}_{2}} R T$

Divide these two equations to get

$\frac{\cancel{P} \cdot {V}_{{H}_{2}}}{\cancel{P} \cdot {V}_{{O}_{2}}} = \frac{{n}_{{H}_{2}} \cdot \cancel{R T}}{{n}_{{O}_{2}} \cdot \cancel{R T}} \implies {V}_{{H}_{2}} / {V}_{{O}_{2}} = {n}_{{H}_{2}} / {n}_{{O}_{2}}$

So, if you need a volume of hydrogen that's twice as big as the volume of oxygen, then your values confirm that both gases will be completely consumed by the reaction.

In other words, $\textcolor{red}{2} {\text{dm}}^{3}$ of ${H}_{2}$ will react with ${\text{1 dm}}^{3}$ of ${O}_{2}$ $\to$ no gas will remain.