Question

Calculate the mass of iron in $10\mathrm{kg}$ of iron ore which contains $80\%$ of pure ferric oxide.

Answer 1

Mass of pure ferric oxide, ${\mathrm{Fe}}_2{\mathrm{O}}_3$ in

$10\mathrm{kg}$ of ore $$\begin{gathered} =\frac{10\times 80}{100} \\ =8\mathrm{kg} \end{gathered}$$

${\mathrm{Fe}}_2{\mathrm{O}}_3=2\mathrm{Fe}$

$$\begin{gathered} 2\times 56+3\times 16=112 \\ 112+48=112 \\ 160\mathrm{g}=112\mathrm{g} \end{gathered}$$

$160\mathrm{g}$ of ${\mathrm{Fe}}_2{\mathrm{O}}_3$ contains $\mathrm{Fe}=112\mathrm{g}$

$8\mathrm{kg}$ of ${\mathrm{Fe}}_2{\mathrm{O}}_3$ contains $\mathrm{Fe}=\frac{112}{160}\times 8$

$=5.6\mathrm{kg}$