Answer
(a) $4 Fe + 3 O_2 -- \gt 2Fe_2O_3$
(b) $ 20.33 mol (Fe)$
(c) $ 1135$ $g (Fe)$
Work Step by Step
(a)
1. Write the reaction, and balance it:
$Fe + O_2 -- \gt Fe_2O_3$
- Balance the $Fe$:
The subscript in the products is "2" from $Fe_2O_3$, so, use that as the coefficient of $Fe$.
$2Fe + O_2 -- \gt Fe_2O_3$
- Balance the $O$:
The subscript in $Fe_2O_3$ for oxygen is "$3$". Therefore, we have to put a "$\frac{3}{2}$" as the coefficient of $O_2$.
* We divided that by 2, because each oxygen molecule has 2 oxygen atoms.
$2Fe + \frac{3}{2}O_2 -- \gt Fe_2O_3$
- Multiply all the coefficients by 2:
$4 Fe + 3 O_2 -- \gt 2Fe_2O_3$
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(b) Each 3 moles of oxygens react with 4 moles of iron. Use that as a conversion factor:
$15.25 mol (O_2) \times \frac{4mol(Fe)}{3mol(O_2)} = 20.33 mol (Fe)$
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(c) Convert the calculated number of moles (Fe) to mass (g):
Molar mass (Fe) = 55.85 g/mol
$20.33 mol(Fe) \times \frac{55.85g(Fe)}{1mol(Fe)} = 1135g(Fe)$
