## Chemistry (7th Edition)

(a) $2Fe_2O_3 + 3C -- \gt 4Fe + 3CO_2$ (b) 4.93 moles of carbon (c) 59.2 g of carbon.
(a) 1. Balance the (O): $2Fe_2O_3 + C -- \gt Fe + 3CO_2$ 2. Balance the (Fe): $2Fe_2O_3 + C -- \gt 4Fe + 3CO_2$ 3. Balance the (C) $2Fe_2O_3 + 3C -- \gt 4Fe + 3CO_2$ ----- (b) and (c) 1. Calculate the number of moles of $Fe_2O_3$: 55.85* 2 + 16.00* 3 = 159.7g/mol $525g \times \frac{1 mol}{ 159.7g} = 3.29mol (Fe_2O_3)$ - The balanced reaction is: $2Fe_2O_3 + 3C -- \gt 4Fe + 3CO_2$ The ratio of $Fe_2O_3$ to $C$ is 2 to 3: $3.29 mol (Fe_2O_3) \times \frac{ 3 mol(C)}{ 2 mol (Fe_2O_3)} = 4.93mol (C)$ 2. Calculate the mass of $C$: 12.01* 1 = 12.01g/mol $4.93 mol \times \frac{ 12.01 g}{ 1 mol} = 59.2g (C)$