To solve this problem, let's go through the steps systematically:
### Step 1: Determine the molar masses
1. Molar mass of Carbon (C):
[tex]\[
\text{Molar mass of C} = 12.01 \, \text{g/mol}
\][/tex]
2. Molar mass of ferric oxide (Fe2O3):
[tex]\[
\text{Molar mass of Fe2O3} = 159.69 \, \text{g/mol}
\][/tex]
### Step 2: Calculate the moles of Carbon
1. Given mass of Carbon:
[tex]\[
\text{Mass of C} = 13.0 \, \text{g}
\][/tex]
2. Moles of Carbon:
[tex]\[
\text{Moles of C} = \frac{\text{Mass of C}}{\text{Molar mass of C}} = \frac{13.0 \, \text{g}}{12.01 \, \text{g/mol}} \approx 1.082 \, \text{moles}
\][/tex]
### Step 3: Determine the stoichiometric relationship
1. From the given balanced equation:
[tex]\[
2 \, \text{Fe}_2\text{O}_3 (s) + 3 \, \text{C} (s) \xrightarrow{\text{heat}} 4 \, \text{Fe} (s) + 3 \, \text{CO}_2 (g)
\][/tex]
The stoichiometric coefficients tell us:
[tex]\[
\text{2 moles of Fe2O3 react with 3 moles of C}
\][/tex]
Therefore:
[tex]\[
\text{For every 1 mole of Fe2O3, } \frac{3}{2} = 1.5 \text{ moles of C are needed}
\][/tex]
### Step 4: Calculate the moles of Fe2O3 needed
1. Moles of Fe2O3 needed:
[tex]\[
\text{Moles of Fe2O3} = \frac{\text{Moles of C}}{1.5} = \frac{1.082 \, \text{moles}}{1.5} \approx 0.722 \, \text{moles}
\][/tex]
### Step 5: Calculate the mass of Fe2O3 needed
1. Mass of Fe2O3 needed:
[tex]\[
\text{Mass of Fe2O3} = \text{Moles of Fe2O3} \times \text{Molar mass of Fe2O3} = 0.722 \, \text{moles} \times 159.69 \, \text{g/mol} \approx 115.24 \, \text{g}
\][/tex]
### Conclusion
The mass of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex] needed to react with 13.0 g of C is approximately [tex]\( 115.24 \)[/tex] grams.
