Answer
$4.20$ g of $Fe$
Work Step by Step
1. Calculate the molar mass of $Fe_2O_3$:
$Fe: 55.85g * 2= 111.7g $
$O: 16.00g * 3= 48.00g $
111.7g + 48.00g = 159.7g
$ \frac{1 mole (Fe_2O_3)}{ 159.7g (Fe_2O_3)}$ and $ \frac{ 159.7g (Fe_2O_3)}{1 mole (Fe_2O_3)}$
2. The balanced reaction is:
$Fe_2O_3 + 3C --\gt 2Fe + 3CO$
According to the coefficients, the ratio of $Fe_2O_3$ to $Fe$ is 1 to 2:
$ \frac{ 2 moles(Fe)}{ 1 mole (Fe_2O_3)}$ and $ \frac{ 1 mole (Fe_2O_3)}{ 2 moles(Fe)}$
3. Calculate the molar mass for $Fe$:
$Fe: 55.85g$
$ \frac{1 mole (Fe)}{ 55.85g (Fe)}$ and $ \frac{ 55.85g (Fe)}{1 mole (Fe)}$
4. Use the conversion factors to find the mass of $Fe$
$6.00g(Fe_2O_3) \times \frac{1 mole(Fe_2O_3)}{ 159.7g( Fe_2O_3)} \times \frac{ 2 moles(Fe)}{ 1 moles (Fe_2O_3)} \times \frac{ 55.85 g (Fe)}{ 1 mole (Fe)} = 4.20g (Fe)$