Question

An iron ore sample contains Fe₂O₃plus other impurities. A 752-g sample of impure iron ore is heated with excess carbon, producing 453 g of pure iron by the following reaction:

$\mathit{F}{\mathit{e}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}\mathbf{3}\mathit{C}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathit{F}\mathit{e}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}\mathbf{3}\mathit{C}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

What is the mass percent of Fe₂O₃ in the impure iron ore sample? Assume that Fe₂O₃ is the only source of iron and that the reaction is 100% efficient.

Short answer

The mass percent of Fe₂O₃ in the impure iron ore sampleis 86.12%

Step-by-step solution

Definition

The molar mass of a materialis defined as the mass of the substancecontained in one moleof the substance, and it isequal to the total massof the component atoms. It may be stated mathematically as follows:
Molar mass =$\frac{Weighting}{numberofmoles}$

Determining the moles of Fe2O3

The balanced equation is given as,

$F{e}_2{O}_3\left(s\right)+3C\left(s\right)\to 2Fe\left(s\right)+3CO\left(g\right)$

Atomic mass of iron=55.845 g/mol

$$\begin{gathered} =\frac{453}{55.845} \\ =8.111mol \end{gathered}$$

From the given reaction, 2 moles of iron $=\frac{8.111mol}{2}$

= 4.0555moles of Fe₂O₃

Determining the mass percentage of Fe2O3

Molar mass$F{e}_2{O}_3=2\left\{Fe\right\}+3\left\{o\right\}$

$$\begin{gathered} =2\left\{55.845\right\}+3\left\{16\right\} \\ =159.69\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right. \end{gathered}$$

Mass of 4.0555moles of Fe₂o₃ = 647.622g

Mass percent ofFe₂o₃ = $\left(\frac{647.622}{752}\right)100$

= 86.12%

Therefore Mass percent is86.12%