Answer
The percent of $Fe_3O_4$ in the ore equals to $3.885\%$.
Work Step by Step
We can calculate the mass of $Fe_3O_4$ required to obtain $2.11g$ of iron:
$m(Fe_3O_4)=m(Fe)×\frac{Mr(Fe_3O_4)}{3×Ar(Fe)}=2.11g×\frac{232}{3\times 56}=2.91g$
Hence, the percent of $Fe_3O_4$ in the ore equals to:
$α=\frac{2.91g}{75g}=3.885\%$.
