a.
\[n_\mathrm{Fe} = \frac {m_\mathrm{Fe}}{M_\mathrm{Fe}} = \frac {250\mathrm{g}}{55,85\frac {\mathrm{g}}{\mathrm{mol}}} = 4,47627574\mathrm{mol} ≈ 4,48\mathrm{mol}\]
b.
\[\begin{aligned} n_\mathrm{FeCl_3}:n_\mathrm{Fe} &= 1:1 \\ n_\mathrm{FeCl_3} &= n_\mathrm{Fe} = 4,47627574\mathrm{mol} \\ m_\mathrm{FeCl_3} &= M_\mathrm{FeCl_3} \cdot n_\mathrm{FeCl_3} = (55,85+35,45 \cdot 3)\frac {\mathrm{g}}{\mathrm{mol}} \cdot 4,47627574\mathrm{mol} = \\ &= 726,051925\mathrm{g} ≈ 726\mathrm{g} \end{aligned}\]
c.
\[\begin{aligned} \text{relativt utbyte} &= \frac {\text{faktiskt utbyte}}{\text{teoretiskt utbyte}} = \frac {632\mathrm{g}}{726,051925\mathrm{g}} = 0,87046116 ≈ \\ &≈ 87,0\% \end{aligned}\]
