geg.: $m(FeS)=20gm(FeS)=20\backslash ;g$

ges.: $m(Fe),m(S)n(FeS)m(Fe),\; \backslash ;\; m(S)\backslash ;n(FeS)$

I. Reaktionsgleichung:

$Fe+S\to FeSFe+S\backslash ;\backslash rightarrow\backslash ;FeS$

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

$n(Fe)=n(S)\iff \frac{n(Fe)}{n(S)}=\frac{1}{1}=1n(Fe)=n(S)\backslash ;\backslash Leftrightarrow\backslash ;\backslash frac\{n(Fe)\}\{n(S)\}=\backslash frac\{1\}\{1\}=1$

III. Ersetzen durch geeignete Quotienten:

$\frac{m(Fe)}{M(Fe)}=\frac{m(S)}{M(S)}\iff \frac{\frac{m(Fe)}{M(Fe)}}{\frac{m(S)}{M(S)}}=\frac{m(Fe)}{M(Fe)}\cdot \frac{M(S)}{m(S)}=1\iff \frac{m(Fe)}{m(S)}=\frac{M(Fe)}{M(S)}\backslash frac\{m(Fe)\}\{M(Fe)\}=\backslash frac\{m(S)\}\{M(S)\}\backslash ;\backslash Leftrightarrow\backslash ;\backslash frac\{\backslash frac\{m(Fe)\}\{M(Fe)\}\}\{\backslash frac\{m(S)\}\{M(S)\}\}=\backslash frac\{m(Fe)\}\{M(Fe)\}\backslash cdot\backslash frac\{M(S)\}\{m(S)\}=1\backslash ;\backslash Leftrightarrow\backslash ;\backslash frac\{m(Fe)\}\{m(S)\}=\backslash frac\{M(Fe)\}\{M(S)\}\backslash ;$ (*)

IV. Einsetzen der Zahlenwerte und berechnen:

aus Angabe: $m(Fe)+m(S)=20gm(Fe)+m(S)=20\backslash ;g$

Zu beachten: Der Quotient $\frac{M(Fe)}{M(S)}=\frac{\mathrm{55,85}\frac{g}{mol}}{\mathrm{32,06}\frac{g}{mol}}\backslash frac\{M(Fe)\}\{M(S)\}=\backslash frac\{55\{,\}85\backslash ;\backslash frac\{g\}\{mol\}\}\{32\{,\}06\backslash ;\backslash frac\{g\}\{mol\}\}$ ergibt einen Zahlenwert!

(Rechenweg zur Auflösung der Gleichung (*) nach $m(Fe)m(Fe)$):

$\frac{m(Fe)}{20g-m(Fe)}=\frac{\mathrm{55,85}\frac{g}{mol}}{\mathrm{32,06}\frac{g}{mol}}\mathrm{\mid }\cdot (20g-m(Fe))\backslash frac\{m(Fe)\}\{20\backslash ;g-m(Fe)\}=\backslash frac\{55\{,\}85\backslash ;\backslash frac\{g\}\{mol\}\}\{32\{,\}06\backslash ;\backslash frac\{g\}\{mol\}\}\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;|\backslash ;\backslash cdot(20\backslash ;g-m(Fe))$

$m(Fe)=10g\cdot \frac{\mathrm{55,85}\frac{g}{mol}}{\mathrm{32,06}\frac{g}{mol}}-m(Fe)\cdot \frac{\mathrm{55,85}\frac{g}{mol}}{\mathrm{32,06}\frac{g}{mol}}\mathrm{\mid }+(m(Fe)\cdot \frac{\mathrm{55,85}\frac{g}{mol}}{\mathrm{32,06}\frac{g}{mol}})m(Fe)=10\backslash ;g\backslash cdot\; \backslash frac\{55\{,\}85\backslash ;\backslash frac\{g\}\{mol\}\}\{32\{,\}06\backslash ;\backslash frac\{g\}\{mol\}\}-m(Fe)\backslash cdot\; \backslash frac\{55\{,\}85\backslash ;\backslash frac\{g\}\{mol\}\}\{32\{,\}06\backslash ;\backslash frac\{g\}\{mol\}\}\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;|\backslash ;+\backslash left(m(Fe)\backslash cdot\; \backslash frac\{55\{,\}85\backslash ;\backslash frac\{g\}\{mol\}\}\{32\{,\}06\backslash ;\backslash frac\{g\}\{mol\}\}\backslash right)$

$m(Fe)\cdot (\frac{\mathrm{55,85}\frac{g}{mol}}{\mathrm{32,06}\frac{g}{mol}}+1)=20g\cdot \frac{\mathrm{55,85}\frac{g}{mol}}{\mathrm{32,06}\frac{g}{mol}}\mathrm{\mid }:(\frac{\mathrm{55,85}\frac{g}{mol}}{\mathrm{32,06}\frac{g}{mol}}+1)m(Fe)\backslash cdot\backslash left(\backslash frac\{55\{,\}85\backslash ;\{\backslash frac\; g\{mol\}\}\}\{32\{,\}06\backslash ;\{\backslash frac\; g\{mol\}\}\}+1\backslash right)=20\backslash ;g\backslash cdot\backslash frac\{55\{,\}85\backslash ;\{\backslash frac\; g\{mol\}\}\}\{32\{,\}06\backslash ;\{\backslash frac\; g\{mol\}\}\}\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;|\backslash ;:\backslash left(\backslash frac\{55\{,\}85\backslash ;\{\backslash frac\; g\{mol\}\}\}\{32\{,\}06\backslash ;\{\backslash frac\; g\{mol\}\}\}+1\backslash right)$

$m(Fe)=\mathrm{12,71}gm(Fe)=12\{,\}71\; \backslash ;g$

$\Rightarrow m(S)=20g-\mathrm{12,71}g=\mathrm{7,29}g\backslash Rightarrow\; \backslash ;m(S)=20\backslash ;g-12\{,\}71\backslash ;g=7\{,\}29\backslash ;g$

A: Bei der Reaktion von Eisen und Schwefel braucht man 12,71 Gramm Eisen und 7,29 Gramm Schwefel.

$n(Fe)=n(FeS)=\frac{m(Fe)}{M(Fe)}=\frac{\mathrm{12,71}g}{\mathrm{55,85}\frac{g}{mol}}=\mathrm{0,228}moln(Fe)=n(FeS)=\backslash frac\{m(Fe)\}\{M(Fe)\}=\backslash frac\{12\{,\}71\backslash ;g\}\{55\{,\}85\backslash ;\backslash frac\{g\}\{mol\}\}=0\{,\}228\backslash ;mol$ (analog mit $m(S)m(S)$ und $M(S)M(S)$ möglich!)

A: Die Stoffmenge von Eisensulfid beträgt 0,228 mol.