## Chemistry 10th Edition

(a) $4 Fe + 3 O_2 -- \gt 2Fe_2O_3$ (b) $20.33 mol (Fe)$ (c) $1135$ $g (Fe)$
(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: $4 Fe + 3 O_2 -- \gt 2Fe_2O_3$ ----------- (b) Each 3 moles of oxygens react with 4 moles of iron. Use that as a conversion factor: $15.25 mol (O_2) \times \frac{4mol(Fe)}{3mol(O_2)} = 20.33 mol (Fe)$ ---------------------------------------------------------- (c) Convert the calculated number of moles (Fe) to mass (g): Molar mass (Fe) = 55.85 g/mol $20.33 mol(Fe) \times \frac{55.85g(Fe)}{1mol(Fe)} = 1135g(Fe)$