# Question 6922c

May 24, 2017

$8.55 {\text{L NH}}_{3}$

#### Explanation:

Let's start by writing the equation for this reaction:

$\text{N"_2 "(g)" + 3"H"_2 "(g)" rightleftharpoons 2"NH"_3 "(g)}$

What we need to do is calculate the moles of hydrogen gas reacting from the given conditions using the ideal-gas equation:

$P V = n R T$

Since we're trying to find the moles, let's rearrange this equation for $n$:

$n = \frac{P V}{R T}$

In order to use the ideal-gas equation, each component must be in the correct units ($\text{L", "atm", "mol}$, and $\text{K}$ for volume, pressure, quantity, and temperature, respectively). We need to convert the temperature to Kelvin and the pressure to atmospheres, and we'll use the conversion factor $\left(1 \text{atm")/(101.325 "kPa}\right)$ and the equation $\text{K} {=}^{o} C + 273$.

The temperature is

$\text{K" = 93.0 ^oC + 273 = 366 "K}$

and the pressure

49.9 cancel("kPa")((1 "atm")/(101.325 cancel("kPa"))) = 0.492 "atm"

and we're given $12.8 \text{L}$.

We can now use the ideal-gas equation to solve for the number of moles of ${\text{H}}_{2}$:

n_("H"_2) = ((0.492 cancel("atm"))(12.8 cancel("L")))/((0.08206 (cancel("L") - cancel("atm"))/("mol"- cancel("K")))(366cancel("K"))) = 0.210 "mol H"_2

Now we can use the stoichiometric relationships (the coefficients) in the chemical equation to calculate the moles of ammonia that can form (assuming the reaction goes to completion):

0.210 cancel("mol H"_2)((2 "mol NH"_3)/(3 cancel("mol H"_2))) = 0.140 "mol NH"_3

We're asked to find the volume of ${\text{NH}}_{3}$ that forms from the given conditions, so we'll again use the ideal-gas equation, but this time to solve for $V$:

$V = \frac{n R T}{P}$

because we known $n$, $T$, and $P$, but not the volume $V$.

Plugging in known variables, the volume of ${\text{NH}}_{3}$ that forms will be

V = ((0.140 "mol")(0.08206 ("L" - cancel("atm"))/(cancel("mol")- cancel("K")))(366 "K"))/(0.492 "atm") = color(red)(8.55 "L"#