# Question a9d2d

May 31, 2017
1. ${\text{H}}_{2}$ is limiting

2. ${\text{N}}_{2}$ is excess

3. $1.3$ ${\text{mol N}}_{2}$ remaining

4. $23 {\text{g NH}}_{3}$ form

#### Explanation:

We're asked to find (1) the limiting reactant in the reaction, (2) the excess reactant, (3) the number of moles of excess reactant remaining after the reaction goes to completion, and (4) the mass of the ammonia gas formed.

1.

First things first, let's write the chemical equation for this reaction:

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

To find the limiting reactant, we must first convert the given quantities of reactants to moles, and then divide by the coefficient in front of it. Whichever quantity after that is lower, that reactant is limiting.

To convert the mass of ${\text{N}}_{2}$ to grams we can use the molar mass of it, which is $2 \times 14.01 \text{g" = 28.02"g"/"mol}$:

56cancel("g N"_2)((1"mol N"_2)/(28.02cancel("g N"_2))) = color(red)(2.0 "mol N"_2

To find the number of moles of ${\text{H}}_{2}$, we can use the fact that at standard temperature and pressure, one mole of any gas occupies a volume of $22.4 \text{L}$, so the moles of ${\text{H}}_{2}$ is

44.8cancel("L H"_2)((1"mol H"_2)/(22.4cancel("L H"_2))) = color(green)(2.0 "mol H"_2

Now let's divide these numbers by their respective coefficients:

${\text{N"_2: color(red)(2.0"mol N"_2)/(1) = 2.0"mol N}}_{2}$

${\text{H"_2: color(green)(2.0"mol H"_2)/(3) = 0.67"mol H}}_{2}$

Since the value for ${\text{H}}_{2}$ is lower, it is the limiting reactant.

2.

Intuitively, since hydrogen is the limiting reactant, ${\text{N}}_{2}$ is the excess reactant.

3.

To find the number of unreacted moles of the excess reactant (${\text{N}}_{2}$), we simply subtract the "used" moles from the original total moles. To find the moles of ${\text{N}}_{2}$ that react, we'll take the quantity of the limiting reactant ($2.0 {\text{mol H}}_{2}$) and use the coefficients:

2.0cancel("mol H"_2)((1"mol N"_2)/(3cancel("mol H"_2))) = color(blue)(0.67"mol N"_2

$2.0 {\text{mol N"_2 - color(blue)(0.67"mol N"_2) = color(orange)(1.3"mol excess N}}_{2}$

4.

Lastly, we need to find the mass of ammonia formed. We can use the coefficients to find the moles of ammonia, and then use its molar mass to find the mass, in $\text{g}$:

2.0cancel("mol H"_2)((2cancel("mol NH"_3))/(3cancel("mol H"_2)))((17.04"g NH"_3)/(1cancel("mol NH"_3))) = color(purple)(23"g NH"_3#