# Question 67fff

Dec 9, 2015

${V}_{N {H}_{3}} = \text{16.7 L}$

#### Explanation:

You don't need to use the ideal gas law equation, you can just use the volume ratio that exists between the gases that take part in the reaction.

The idea is that when gases react under the same conditions for pressure and temperature, the mole ratios that exist between the reactants and products are equivalent to volume ratios.

A quick demonstration of why that it is so. Let's assume that this reaction takes place at a pressure $P$ and a temperature $T$ for all three gases.

This means that you can use the ideal gas law equation to write

$P \cdot {V}_{{H}_{2}} = {n}_{{H}_{2}} \cdot R \cdot T \to$ for hydrogen gas

$P \cdot {V}_{{N}_{2}} = {n}_{{N}_{2}} \cdot R \cdot T \to$ for nitrogen gas

Divide these equations to get

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

This is equivalent to

${\overbrace{{n}_{{H}_{2}} / {n}_{{N}_{2}}}}^{\textcolor{b l u e}{\text{mole ratio")) = overbrace(V_(H_2)/V_(N_2))^(color(purple)("volume ratio}}}$

As you can see, the mole ratio is equivalent to the volume ratio for any pair of gases that take part in the reaction.

In your case, the balanced chemical equation for this synthesis reaction looks like this

$\textcolor{red}{3} {\text{H"_text(2(g]) + "N"_text(2(g]) -> color(blue)(2)"NH}}_{\textrm{3 \left(g\right]}}$

Notice that you have a $\textcolor{red}{3} : 1$ mole ratio between hydrogen gas and nitrogen gas. This of course means that you also have a $\textcolor{red}{3} : 1$ volume ratio between these two gases.

Use the given volumes to see if one of the reactants acts as a limiting reagent. To do that, pick one reactant and use the volume ratio to see if you have enough of the second reactant

25.0 color(red)(cancel(color(black)("L H"_2))) * "1 L N"_2/(color(red)(3)color(red)(cancel(color(black)("L H"_2)))) = "8.333 L N"_2

Since you have $\text{15.0 L}$ of nitrogen gas available, you can say that nitrogen will be in excess, or, in other words, hydrogen gas will be the limiting reagent.

Hydrogen gas will be completely consumed by the reaction, and you'll be left with an excess of $\text{6.667 L}$ of nitrogen gas.

Now look at the $\textcolor{red}{3} : \textcolor{b l u e}{2}$ mole ratio that exists between hydrogen and ammonia. If $\text{25.0 L}$ of hydrogen gas react, you can say that the reaction will produce

25.0 color(red)(cancel(color(black)("L H"_2))) * (color(blue)(2)" moles NH"_3)/(color(red)(3) color(red)(cancel(color(black)("L H"_2)))) = "16.667 L NH"_3#

Rounded to three sig figs, the answer will be

${V}_{N {H}_{3}} = \textcolor{g r e e n}{\text{16.7 L}}$

Similar problems

https://socratic.org/questions/how-many-liters-of-o2-g-are-needed-to-react-completely-with-56-0-l-of-ch4-g-at-s

https://socratic.org/questions/at-the-same-temperature-and-pressure-what-system-volume-will-be-taken-up-when-10