Question

The reusable booster rockets of the U.S. space shuttle use a mixture of aluminium andammonium per chlorate for fuel. A possible equation for this reaction is
$3Al\left(\mathrm{s}\right)+3N{H}_{4}Cl{O}_4\left(\mathrm{s}\right)\to A{l}_2{\mathrm{O}}_3\left(\mathrm{s}\right)+AlC{l}_3\left(\mathrm{s}\right)+3NO\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$
what mass of$N{H}_{4}Cl{O}_4$should be used in the fuel mixture for every kilogram of?

Short answer

The mass ${\mathrm{NH}}_4{\mathrm{ClO}}_4$of should be used in the fuel mixture for every kilogram of $\mathrm{Al}$is$4353.2\mathrm{g}$

Step-by-step solution

 Step 1: Definition

The mass of any material expressed in atomic mass units is quantitatively equivalent to its molar mass in grams per mole.

Determination of number of moles

The given balanced equation is,

$3\mathrm{Al}\left(\mathrm{s}\right)+3{\mathrm{NH}}_4{\mathrm{ClO}}_4\left(\mathrm{s}\right)\to {\mathrm{Al}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)+{\mathrm{AlCl}}_3\left(\mathrm{s}\right)+3\mathrm{NO}\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

From the above chemical equation, for 3 moles of $\mathrm{Al}$, 3 moles of ${\mathrm{NH}}_4{\mathrm{ClO}}_4$is required

For $1\mathrm{kg}$, Number of moles $=1000\mathrm{gAl}$

$$\begin{gathered} =1000\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$26.98\mathrm{as}\mathrm{Atomic}\mathrm{mass}=26.98$}\right. \\ =37.0644\mathrm{Moles} \\ \mathrm{Thus},37.0644\mathrm{moles}\mathrm{of}\mathrm{aluminium}\mathrm{requries}37.0644\mathrm{moles}\mathrm{of}{\mathrm{NH}}_4{\mathrm{ClO}}_4. \end{gathered}$$

Determination of molar mass of NH4ClO4

Molar mass of

$$\begin{gathered} {\mathrm{NH}}_4{\mathrm{ClO}}_4=\left\{\mathrm{N}\right\}+4\left\{\mathrm{H}\right\}+\left\{\mathrm{C}\right\}+4\left\{\mathrm{O}\right\} \\ =14.00+4.00+35.45+64.00 \\ =117.45\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right. \end{gathered}$$

thus, mass

$\begin{array}{rcl}{\mathrm{NH}}_4{\mathrm{ClO}}_4\mathrm{required}& =& \left(\mathrm{moles}\right)\left(\mathrm{molarance}\right)\\ & =& \left(37.0644\right)\left(117.45\right)\\ \mathrm{Mass}\mathrm{of}{\mathrm{NH}}_4{\mathrm{ClO}}_4\mathrm{required}& =& 4353.2\mathrm{g}\end{array}$