Answer
(a) $2Fe_2O_3 + 3C -- \gt 4Fe + 3CO_2$
(b) 4.93 moles of carbon
(c) 59.2 g of carbon.
Work Step by Step
(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)
$2Fe_2O_3 + 3C -- \gt 4Fe + 3CO_2$
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(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
$4.93 mol \times \frac{ 12.01 g}{ 1 mol} = 59.2g (C)$
