Question

At liftoff, a space shuttle uses a solid mixture of ammonium perchlorate and aluminum powder to obtain great thrust from the volume change of solid to gas. In the presence of a catalyst, the mixture forms solid aluminum oxide and aluminium trichloride and gaseous water and nitrogen monoxide. (a) Write a balanced equation for the reaction, and identify the reducing and oxidizing agents. (b) How many total moles of gas (water vapor and nitrogen monoxide) are produced when 50.0 kg of ammonium perchlorate reacts with a stoichiometric amount of Al? (c) What is the volume change from this reaction? (dof NH₄ClO₄ = 1.95 g/cc, Al = 2.70 g/cc, Al₂O₃ = 3.97 g/cc, and AlCl₃ = 2.44 g/cc; assume 1 mol of gas occupies 22.4 L.)

Short answer

1. The balanced reaction is, $\mathbf{3}{\mathbf{NH}}_{\mathbf{4}}{\mathbf{ClO}}_{\mathbf{4}}\left(s\right)\mathbf{+}\mathbf{3}\mathbf{Al}\left(s\right)\mathbf{\to }{\mathbf{Al}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\left(s\right)\mathbf{+}{\mathbf{AlCl}}_{\mathbf{3}}\left(s\right)\mathbf{+}\mathbf{6}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(g\right)\mathbf{+}\mathbf{3}\mathbf{NO}\left(g\right)$
2. Thus, the moles of H₂O and NO are 851.2 mol and 425.2 mol, respectively.
3. The volume is 2.86x10⁴ L.

Step-by-step solution

(a) Balanced reaction

The balanced equation for the reaction is,

$\mathbf{3}{\mathbf{NH}}_{\mathbf{4}}{\mathbf{ClO}}_{\mathbf{4}}\left(s\right)\mathbf{+}\mathbf{3}\mathbf{Al}\left(s\right)\mathbf{\to }{\mathbf{Al}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\left(s\right)\mathbf{+}{\mathbf{AlCl}}_{\mathbf{3}}\left(s\right)\mathbf{+}\mathbf{6}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(g\right)\mathbf{+}\mathbf{3}\mathbf{NO}\left(g\right)$

In this reaction, the oxidation state of Al changes from 0 to +3.

Thus, Al act as reducing agent and NH₄ClO₄ act as oxidising agent.

(b) Determination of moles of nitrogen oxide and water

Moles of NH₄ClO₄ are,

$\begin{array}{rcl}\mathbf{Moles}& \mathbf{=}& \frac{\mathbf{mass}}{\mathbf{molar}\mathbf{}\mathbf{mass}}\\ & \mathbf{=}& \frac{\mathbf{50000}\mathbf{g}}{\mathbf{117}\mathbf{.}\mathbf{49}\mathbf{g}\mathbf{/}\mathbf{mol}}\\ & \mathbf{=}& \mathbf{425}\mathbf{.}\mathbf{6}\mathbf{mol}\end{array}$

From the reaction, it can be concluded that 3 mol of NH₄ClO₄ produces 6 mol of H₂O and 3 mol of NO.

Moles of H­₂O are,

$$\begin{gathered} \mathbf{=}\frac{\mathbf{425}\mathbf{.}\mathbf{6}\mathbf{mol}\mathbf{\times }\mathbf{6}\mathbf{mol}}{\mathbf{3}\mathbf{mol}} \\ \mathbf{=}\mathbf{851}\mathbf{.}\mathbf{2}\mathbf{mol} \end{gathered}$$

Moles of NO are,

$$\begin{gathered} \mathbf{=}\frac{\mathbf{425}\mathbf{.}\mathbf{6}\mathbf{mol}\mathbf{\times }\mathbf{3}\mathbf{mol}}{\mathbf{3}\mathbf{mol}} \\ \mathbf{=}\mathbf{425}\mathbf{.}\mathbf{6}\mathbf{mol} \end{gathered}$$

Thus, the moles of H₂O and NO are 851.2 mol and 425.2 mol, respectively.

(c) Determination of change in volume

Total moles of gas are,

$$\begin{gathered} \mathbf{=}\mathbf{851}\mathbf{.}\mathbf{2}\mathbf{mol}\mathbf{+}\mathbf{425}\mathbf{.}\mathbf{6}\mathbf{mol} \\ \mathbf{=}\mathbf{1276}\mathbf{.}\mathbf{8}\mathbf{mol} \end{gathered}$$

Volume of the gas is,

$\begin{array}{rcl}\mathbf{1}\mathbf{mol}& \mathbf{=}& \mathbf{22}\mathbf{.}\mathbf{4}\mathbf{L}\\ \mathbf{1276}\mathbf{.}\mathbf{8}\mathbf{mol}& \mathbf{=}& \mathbf{2}\mathbf{.}\mathbf{86}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{L}\end{array}$

Thus, the volume is 2.86x10⁴ L.