# What volume of ammonia is produced by the reaction when "19.5 L" of hydrogen gas react at 93^@"C" and "48.7 kPa" ?

Feb 18, 2017

$\text{13.0 L}$

#### Explanation:

The trick here is to realize that all the chemical species that take part in this reaction are kept under the same conditions for pressure and temperature.

This means that the mole ratios that exists between the chemical species that take part in the reaction will be equivalent to volume ratios.

The balanced chemical equation that describes this reaction looks like this

${\text{N"_ (2(g)) + color(blue)(3)"H"_ (2(g)) -> color(darkorange)(2)"NH}}_{3 \left(g\right)}$

Notice that for every $\textcolor{b l u e}{3}$ moles of hydrogen gas that take part in the reaction, the reaction produces $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{2}$ moles of ammonia.

When the reaction takes place at constant temperature and pressure, you can say that for every $\textcolor{b l u e}{3}$ liters of hydrogen gas that react, the reaction produces $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{2}$ liters of ammonia.

Assuming that nitrogen gas is in excess, you can say that the given volume of hydrogen gas will produce

$19.5 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{L H"_2))) * (color(darkorange)(2)color(white)(.)"moles NH"_3)/(color(blue)(3)color(red)(cancel(color(black)("L H"_2)))) = color(darkgreen)(ul(color(black)("13.0 L NH}}_{3}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the volume of hydrogen gas.

Keep in mind that this volume of ammonia is produced at a temperature of ${93.0}^{\circ} \text{C}$ and a pressure of $\text{48.7 kPa}$.