Answer:
Approximately [tex]0.11823\; \rm g[/tex] in total for the five tablets.
(Approximately [tex]2.3646 \times 10^{-2}\; \rm g[/tex] per tablet.)
Explanation:
The relative atomic mass of an element is the ratio between:
- the mass of one atom of this element, and
- the mass of one-twelfth of a carbon-[tex]12[/tex] atom.
The definition of moles ensures that the relative atomic mass of an element is approximately numerically equal to the mass (measured in grams) of one mole of the atoms of this element.
The question states that the relative atomic mass [tex]A_{\rm r}[/tex] of iron is [tex]56[/tex]. In other words, the mass of one mole of iron [tex]\rm Fe[/tex] atoms would be approximately [tex]56\; \rm g[/tex].
The question is asking for the mass of some amount of [tex]\rm Fe^{2+}[/tex] ions. Each [tex]\rm Fe^{2+}\![/tex] ion contains two fewer electrons than a neutral [tex]\rm Fe[/tex] atom. Hence, [tex]4.2225 \times 10^{4}\; \rm mol[/tex] of [tex]\rm Fe^{2+} \![/tex] ions might be lighter than the same number of [tex]\rm Fe \![/tex] atoms by a very small extent: The mass of one mole of electrons is approximately [tex]5 \times 10^{-4}\; \rm g[/tex], much smaller than the mass of the same number of [tex]\!\rm Fe[/tex] atoms (approximately [tex]56\; \rm g[/tex].)
Estimate the mass of these [tex]4.2225 \times 10^{4}\; \rm mol[/tex] of [tex]\rm Fe^{2+}[/tex] ions using the mass of the same number of [tex]\rm Fe[/tex] atoms:
[tex]\begin{aligned}&m({\rm Fe^{2+}}) \\ &\approx m({\rm Fe}) \\ &\approx 4.2225 \times 10^{-4}\; \rm mol \\ &\quad\quad \times 56\; \rm g \cdot mol^{-1} \\ &\approx 2.3646 \times 10^{-2}\; \rm g \end{aligned}[/tex].
Five of these tablets would contain approximately [tex]5 \times 2.3646 \times 10^{-2}\; \rm g \approx 0.11823\; \rm g[/tex] of [tex]\rm Fe^{2+}[/tex] ions.
