Answer:
The reaction equation is given as:
\[ \text{Fe}_2\text{O}_3 + 3\text{CO} \to 2\text{Fe} + 3\text{CO}_2 \]
First, we need to determine the molar mass of iron(III) oxide (\(\text{Fe}_2\text{O}_3\)). The molar masses of Fe and O are approximately 55.85 g/mol and 16.00 g/mol, respectively.
\[ \text{Molar mass of } \text{Fe}_2\text{O}_3 = 2(55.85) + 3(16.00) = 111.70 + 48.00 = 159.70 \, \text{g/mol} \]
Next, let's calculate the number of moles in 10 g of \(\text{Fe}_2\text{O}_3\):
\[ \text{Moles of } \text{Fe}_2\text{O}_3 = \frac{10 \, \text{g}}{159.70 \, \text{g/mol}} = 0.0626 \, \text{mol} \]
The balanced reaction equation tells us that 1 mole of \(\text{Fe}_2\text{O}_3\) will produce 2 moles of Fe. So, using the moles of \(\text{Fe}_2\text{O}_3\) calculated, we can find the moles of Fe that will be produced:
\[ \text{Moles of Fe} = 0.0626 \, \text{mol} \times 2 = 0.1252 \, \text{mol} \]
Next, we need to determine the molar mass of Fe, which is 55.85 g/mol. Now, let's calculate the mass of Fe produced:
\[ \text{Mass of Fe} = 0.1252 \, \text{mol} \times 55.85 \, \text{g/mol} = 6.99 \, \text{g} \]
Therefore, the mass of Fe that will be produced from 10 grams of iron(iii) oxide from the reaction is approximately 6.99 grams.
