Question

Twenty dietary iron tablets with a total mass of 22.131 g

were ground and mixed thoroughly. Then 2.998 g of the powder were dissolved in${\mathbf{HNO}}_{\mathbf{3}}$and heated to convert all iron into${\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}$Addition of${\mathbf{NH}}_{\mathbf{3}}$precipitated${\mathbf{Fe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{\times }{\mathbf{xH}}_{\mathbf{2}}\mathbf{O}$, which was ignited to give$\mathbf{0}\mathbf{.}\mathbf{264}{\mathbf{gofFe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{(}\mathbf{FM}\mathbf{159}\mathbf{.}\mathbf{69}\mathbf{)}$.What is the average mass of${\mathbf{FeSO}}_{\mathbf{4}}\mathbf{\times }\mathbf{7}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{(}\mathbf{FM}\mathbf{278}\mathbf{.}\mathbf{01}\mathbf{)}$in each tablet?

Short answer

The value of mass of 1 tablet is$6.786g/20=0.339g$

Step-by-step solution

Calculate n(Fe):

$\mathrm{n}\left({\mathrm{Fe}}_2{\mathrm{O}}_3\right)=\mathrm{n}\left(\mathrm{Fe}\right)$

1 mol of$\left({\mathrm{Fe}}_2{\mathrm{O}}_3\right)$ contains 2 mole of Fe.

$$\begin{gathered} \mathrm{n}\left(\mathrm{Fe}\right)=2\times \frac{\mathrm{m}\left({\mathrm{Fe}}_2{\mathrm{O}}_3\right)}{\mathrm{M}\left({\mathrm{Fe}}_2{\mathrm{O}}_3\right)} \\ \mathrm{n}\left(\mathrm{Fe}\right)=\frac{2\times 0.264\mathrm{g}}{159.69\mathrm{g}/\mathrm{mol}} \end{gathered}$$

The moles of n(Fe) is$\mathrm{n}\left(\mathrm{Fe}\right)=3.306\times {10}^{-3}\mathrm{mol}$

Calculate the mass of 1 tablet:

Calculate${\mathrm{m}}_1({\mathrm{FeSO}}_4\times 7{\mathrm{H}}_2\mathrm{O})$,

$$\begin{gathered} \mathrm{n}\left(\mathrm{Fe}\right)=\mathrm{n}({\mathrm{FeSO}}_4\times 7{\mathrm{H}}_2\mathrm{O}) \\ {\mathrm{m}}_1({\mathrm{FeSO}}_4\times 7{\mathrm{H}}_2\mathrm{O})=\mathrm{n}\times \mathrm{M} \\ {\mathrm{m}}_1({\mathrm{FeSO}}_4\times 7{\mathrm{H}}_2\mathrm{O})=(3.306\times {10}^{-3}\mathrm{mol})\times 278.01\mathrm{g}/\mathrm{mol} \\ {\mathrm{m}}_1({\mathrm{FeSO}}_4\times 7{\mathrm{H}}_2\mathrm{O})=0.9192\mathrm{g} \end{gathered}$$

Calculate${\mathrm{m}}_2({\mathrm{FeSO}}_4\times 7{\mathrm{H}}_2\mathrm{O})$,

$$\begin{gathered} {\mathrm{m}}_2({\mathrm{FeSO}}_4\times 7{\mathrm{H}}_2\mathrm{O})=\frac{{\mathrm{m}}_{\mathrm{total}}}{{\mathrm{m}}_{\mathrm{analyzed}}}\times {\mathrm{m}}_1 \\ {\mathrm{m}}_2({\mathrm{FeSO}}_4\times 7{\mathrm{H}}_2\mathrm{O})=\frac{22.131\mathrm{g}}{2.998\mathrm{g}\times 0.9192\mathrm{g}} \\ {\mathrm{m}}_2({\mathrm{FeSO}}_4\times 7{\mathrm{H}}_2\mathrm{O})=6.786\mathrm{g} \end{gathered}$$

Then mass of 1 tablet is$6.786\mathrm{g}/20$$=0.339g$