Answer
$84.0$ g of $CO$ are produced.
Work Step by Step
1. Calculate the molar mass of $C$:
$C: 12.01g$
$ \frac{1 mole (C)}{ 12.01g (C)}$ and $ \frac{ 12.01g (C)}{1 mole (C)}$
2. The balanced reaction is:
$Fe_2O_3 + 3C --\gt 2Fe + 3CO$
According to the coefficients, the ratio of $C$ to $CO$ is 3 to 3:
$ \frac{ 3 moles(CO)}{ 3 moles (C)}$ and $ \frac{ 3 moles (C)}{ 3 moles(CO)}$
3. Calculate the molar mass for $CO$:
$C: 12.01g $
$O: 16g$
12.01g + 16.00g = 28.01g
$ \frac{1 mole (CO)}{ 28.01g (CO)}$ and $ \frac{ 28.01g (CO)}{1 mole (CO)}$
4. Use the conversion factors to find the mass of $CO$
$36.0g(C) \times \frac{1 mole(C)}{ 12.01g( C)} \times \frac{ 3 moles(CO)}{ 3 moles (C)} \times \frac{ 28.01 g (CO)}{ 1 mole (CO)} = 84.0g (CO)$
