Answer
There are 73.2 kg of $Fe$ in $Fe_2O_3$
Work Step by Step
105 kg = 105 $\times 10^3$ g
1. Calculate the number of moles of $Fe_2O_3$:
55.85* 2 + 16* 3 = 159.7g/mol
$105 \times 10^3g \times \frac{1 mol}{ 159.7g} = 0.657 \times 10^3 mol (Fe_2O_3)$
Each $Fe_2O_3$ has 2 $Fe$, so:
The ratio of $Fe_2O_3$ to $Fe$ is 1 to 2:
$0.657 mol \times 10^3 (Fe_2O_3) \times \frac{ 2 mol(Fe)}{ 1 mol (Fe_2O_3)} = 1.31 \times 10^3 mol (Fe)$
2. Calculate the mass of $Fe$:
55.85* 1 = 55.85g/mol
$1.31 \times 10^3 mol \times \frac{ 55.85 g}{ 1 mol} = 73.2 \times 10^3g (Fe)$ = 73.2 kg
