Question

A mixture containing only ${\mathbf{Al}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$(FM 101.96) and${\mathbf{Fe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$(FM 159.69) weighs 2.019 g. When heated under a stream of${\mathbf{H}}_{\mathbf{2}\mathbf{}\mathbf{}}{\mathbf{Al}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$is unchanged, butis${\mathbf{Fe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$ converted into metallic Fe plus${\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$If the residue weighs 1.774 g, what is the weight percent of${\mathbf{Fe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$in the original mixture?

Short answer

The weight percent of ${\mathrm{Fe}}_2{\mathrm{O}}_3$is 0.404%.

Step-by-step solution

Calculate the weight percent:

From the question we can consider the reaction as,

${\mathrm{Fe}}_2{\mathrm{O}}_3+{\mathrm{Al}}_2{\mathrm{O}}_3+\mathrm{heat}\to \mathrm{Fe}+{\mathrm{Al}}_2{\mathrm{O}}_3$

Consider,

localid="1663665835168" ${\mathrm{m}}_1({\mathrm{Fe}}_2{\mathrm{O}}_3+{\mathrm{Al}}_2{\mathrm{O}}_3)$as 2.019g,

${\mathrm{m}}_2(\mathrm{Fe}+{\mathrm{Al}}_2{\mathrm{O}}_3)$as 1.774g

Mass of lost oxygen is $\mathrm{m}\left(\mathrm{O}\right)={\mathrm{m}}_1-{\mathrm{m}}_2$

$$\begin{gathered} \mathrm{m}\left(\mathrm{O}\right)=2.019\mathrm{g}-1.774\mathrm{g} \\ \mathrm{m}\left(\mathrm{O}\right)=0.245\mathrm{g} \\ 0.245\mathrm{g}ofoxygen=0.01531moles \end{gathered}$$

Calculate the moles of localid="1663665958980" ${\mathrm{Fe}}_2{\mathrm{O}}_3$

$$\begin{gathered} \mathrm{n}\left({\mathrm{Fe}}_2{\mathrm{O}}_3\right)=\frac{1}{3}\times \mathrm{n}\left(\mathrm{O}\right) \\ \mathrm{n}\left({\mathrm{Fe}}_2{\mathrm{O}}_3\right)=\frac{1}{3}\times 0.01531\mathrm{mol} \\ \mathrm{n}\left({\mathrm{Fe}}_2{\mathrm{O}}_3\right)=0.005105\mathrm{mol} \end{gathered}$$

So the weight percentage is

$$\begin{gathered} wt\%=\frac{m\left(F{e}_2{O}_3\right)}{m\left(sample\right)} \\ wt\%=\frac{0.815g}{2.019g} \\ wt\%=0.404 \end{gathered}$$

The weight percent of $F{e}_2{O}_3$is 0.404%.