| Cr2O3 | + 2Al | → Al2O3 | + 2Cr | |
| \[m\] | ①\[2,5\mathrm{ton}\] | ④\[1 710 526,32\mathrm{g}\] | ||
| \[n\] | ②\[16 447,3684\mathrm{mol}\] | ③\[32 894,7378\mathrm{mol}\] |
① \(m_\mathrm{Cr_2O_3} = 2,5\mathrm{ton} = 2,5 \cdot 10^6\mathrm{g}\)
②
\(\begin{aligned}n_\mathrm{Cr_2O_3} = \frac {m_\mathrm{Cr_2O_3}}{M_\mathrm{Cr_2O_3}} = \frac {2,5 \cdot 10^6\mathrm{g}}{(52,00 \cdot 2 + 16,00 \cdot 3)\mathrm{g/mol}} = 16447,3684\mathrm{mol}\end{aligned}\)
③
\(n_\mathrm{Cr_2O_3}:n_\mathrm{Cr} = 1:2 \\ n_\mathrm{Cr} = 2n_\mathrm{Cr_2O_3} = 2 \cdot 16447,3684\mathrm{mol} = 32894,7368\mathrm{mol}\)
④
\(\begin{aligned}m_\mathrm{Cr} &= M_\mathrm{Cr} \cdot n_\mathrm{Cr} = 52,00\mathrm{g/mol} \cdot 32894,7368\mathrm{mol} = \\ &= 1710526,32\mathrm{g} = \text{teoretiskt utbyte}\end{aligned}\)
Vi beräknar det faktiska utbytet:
\[\begin{aligned}\text{relativt utbyte} &= \frac {\text{faktiskt utbyte}}{\text{teoretiskt utbyte}} \\ \text{faktiskt utbyte} &= \text{relativt utbyte} \cdot \text{teoretiskt utbyte} = 0,92 \cdot 1710526,32\mathrm{g} = \\ &= 1573684,21\mathrm{g} = 1,57368421 \cdot 10^6\mathrm{g} ≈ 1,6\mathrm{ton}\end{aligned}\]
