Answer
7.98 g of $Fe_2O_3$
Work Step by Step
1. Calculate the number of moles of $Fe$:
55.85* 1 = 55.85g/mol
$5.58g \times \frac{1 mol}{ 55.85g} = 0.100mol (Fe)$
- As we determined in the last exercise, the balanced reaction is:
$4Fe + 3O_2 -- \gt 2Fe_2O_3$
The ratio of $Fe$ to $Fe_2O_3$ is 4 to 2:
$0.100 mol (Fe) \times \frac{ 2 mol(Fe_2O_3)}{ 4 mol (Fe)} = 0.050mol (Fe_2O_3)$
2. Calculate the mass of $Fe_2O_3$:
55.85* 2 + 16* 3 = 159.7g/mol
$0.050 mol \times \frac{ 159.7 g}{ 1 mol} = 7.98g (Fe_2O_3)$
