Answer
We can obtain 699g of $Fe$, but, the percent yield was only 93.5%.
Work Step by Step
1. Calculate the number of moles of $Fe_2O_3$:
55.85* 2 + 16* 3 = 159.7g/mol
$1000g \times \frac{1 mol}{ 159.7g} = 6.26mol (Fe_2O_3)$
- The balanced reaction is:
$Fe_2O_3 + 3CO -- \gt 2Fe + 3CO_2$
The ratio of $Fe_2O_3$ to $Fe$ is 1 to 2:
$6.26 mol (Fe_2O_3) \times \frac{ 2 mol(Fe)}{ 1 mol (Fe_2O_3)} = 12.5mol (Fe)$
2. Calculate the mass of $Fe$:
55.85* 1 = 55.85g/mol
$12.5 mol \times \frac{ 55.85 g}{ 1 mol} = 699g (Fe)$
3. Now, calculate the percent yield:
$\frac{ 654g}{699g} \times 100\% = 93.5\%$
