3MnO2 + 4Al → 2Al2O3 + 3Mn
\[n_\mathrm{MnO_2} = \frac {m_\mathrm{MnO_2}}{M_\mathrm{MnO_2}} = \frac {63,8\mathrm{g}}{(54,94+16,00 \cdot 2)\frac {\mathrm{g}}{\mathrm{mol}}} = 0,73383943\mathrm{mol}\]\[n_\mathrm{Al} = \frac {m_\mathrm{Al}}{M_\mathrm{Al}}= \frac {25,5\mathrm{g}}{26,98\frac {\mathrm{g}}{\mathrm{mol}}}= 0,94514455\mathrm{mol}\]\[\frac {n_\mathrm{Al}}{n_\mathrm{MnO_2}} = \frac {0,94514455\mathrm{mol}}{0,73383943\mathrm{mol}}= 1,28794463< \frac {4}{3}\]
\(\Rightarrow n_\mathrm{Al}\) är begränsande.
| 3MnO2 | + 4Al | → 2Al2O3 | + 3Mn | |
| \[m\] | ①\[25,5\mathrm{g}\] | ④\[38,9\mathrm{g}\] | ||
| \[n\] | ②\[0,94514455\mathrm{mol}\] | ③\[0,70885841\mathrm{mol}\] |
① \(m_\mathrm{Al} = 25,5\mathrm{g}\)
② \(n_\mathrm{Al}= 0,94514455\mathrm{mol}\)
③ \(n_\mathrm{Al}:n_\mathrm{Mn} = 4:3 \\ n_\mathrm{Mn} = \frac {3}{4} n_\mathrm{Al} = \frac {3}{4} \cdot 0,94514455\mathrm{mol} = 0,70885841\mathrm{mol}\)
④ \(m_\mathrm{Mn} = M_\mathrm{Mn} \cdot n_\mathrm{Mn} = 54,94\mathrm{g/mol} \cdot 0,70885841\mathrm{mol} = 38,9446812\mathrm{g} ≈ 38,9\mathrm{g}\)
Svar: Det kan bildas maximalt 38,9 g mangan.
