geg.: $m(Fe)=500gm(Fe)=500\backslash ;g$

ges.: $m(F{e}_2{O}_3)m(Fe\_2O\_3)$

I. Reaktionsgleichung:

$F{e}_2{O}_3+2Al\to 2Fe+A{l}_2{O}_{3}Fe\_2O\_3+2\backslash ;Al\backslash rightarrow\backslash ;2\backslash ;Fe+Al\_2O\_3$

II. Bestimmen des Stoffmengenverhältnisses (vgl. Koeffizienten der Reaktionsgleichung):

$\frac{n(F{e}_2{O}_3)}{n(Fe)}=\frac{1}{2}\iff n(F{e}_2{O}_3)=\frac{1}{2}\cdot n(Fe)\backslash frac\{n(Fe\_2O\_3)\}\{n(Fe)\}=\backslash frac\{1\}\{2\}\backslash ;\backslash Leftrightarrow\backslash ;n(Fe\_2O\_3)=\backslash frac\{1\}\{2\}\backslash cdot\; n(Fe)$

III. Ersetzen durch geeignete Quotienten:

$\frac{m(F{e}_2{O}_3)}{M(F{e}_2{O}_3)}=\frac{1}{2}\cdot \frac{m(Fe)}{M(Fe)}\iff m(F{e}_2{O}_3)=\frac{1}{2}\cdot \frac{m(Fe)}{M(Fe)}\cdot M(F{e}_2{O}_3)\backslash frac\{m(Fe\_2O\_3)\}\{M(Fe\_2O\_3)\}=\backslash frac\{1\}\{2\}\backslash cdot\; \backslash frac\{m(Fe)\}\{M(Fe)\}\backslash ;\backslash Leftrightarrow\backslash ;m(Fe\_2O\_3)=\backslash frac\{1\}\{2\}\backslash cdot\; \backslash frac\{m(Fe)\}\{M(Fe)\}\backslash cdot\; M(Fe\_2O\_3)$

IV. Einsetzen der Zahlenwerte und berechnen:

$m(F{e}_2{O}_3)=\frac{1}{2}\cdot \frac{500g}{\mathrm{55,8}\frac{g}{mol}}\cdot \mathrm{159,6}\frac{g}{mol}=\mathrm{715,1}gm(Fe\_2O\_3)=\backslash frac\{1\}\{2\}\backslash cdot\; \backslash frac\{500\backslash ;g\}\{55\{,\}8\backslash ;\backslash frac\{g\}\{mol\}\}\backslash cdot\; 159\{,\}6\backslash ;\; \backslash frac\{g\}\{mol\}=\; 715\{,\}1\backslash ;g$

A: Für dieses Thermitverfahren sind 715,1 Gramm Eisen(III)oxid erforderlich.