Answer
2.40 g of $O_2$
Work Step by Step
1. Calculate the number of moles of $Fe$:
55.85* 1 = 55.85g/mol
$5.58g \times \frac{1 mol}{ 55.85g} = 0.100mol (Fe)$
- The balanced reaction is:
$4Fe + 3O_2 -- \gt 2Fe_2O_3$
The ratio of $Fe$ to $O_2$ is 4 to 3:
$0.100 mol (Fe) \times \frac{ 3 mol(O_2)}{ 4 mol (Fe)} = 0.0750mol (O_2)$
2. Calculate the mass of $O_2$:
16.00* 2 = 32.00g/mol
$0.0750 mol \times \frac{ 32.00 g}{ 1 mol} = 2.40g (O_2)$
