To find out how many grams of CO would be required to generate 635 grams of CO[tex]\(_2\)[/tex], we will follow a series of stoichiometric calculations based on the given chemical equation:
[tex]\[
\text{Fe}_2\text{O}_3 + 3 \text{CO} \rightarrow 2 \text{Fe} + 3 \text{CO}_2
\][/tex]
The molar masses provided are:
- [tex]\( \text{CO}_2 \)[/tex]: 44.01 g/mol
- [tex]\( \text{CO} \)[/tex]: 28.01 g/mol
### Step-by-Step Solution
1. Calculate the number of moles of [tex]\( \text{CO}_2 \)[/tex] produced:
First, we need to convert the mass of [tex]\( \text{CO}_2 \)[/tex] into moles. We use the molar mass of [tex]\( \text{CO}_2 \)[/tex]:
[tex]\[
\text{Number of moles of } \text{CO}_2 = \frac{\text{mass of } \text{CO}_2}{\text{molar mass of } \text{CO}_2}
\][/tex]
[tex]\[
\text{Number of moles of } \text{CO}_2 = \frac{635 \text{ g}}{44.01 \text{ g/mol}} \approx 14.43 \text{ moles}
\][/tex]
2. Use stoichiometry to find the number of moles of [tex]\( \text{CO} \)[/tex] required:
According to the balanced chemical equation, 3 moles of [tex]\( \text{CO} \)[/tex] produce 3 moles of [tex]\( \text{CO}_2 \)[/tex]. This means that the mole ratio of [tex]\( \text{CO} \)[/tex] to [tex]\( \text{CO}_2 \)[/tex] is 1:1.
Therefore, the number of moles of [tex]\( \text{CO} \)[/tex] required to produce 14.43 moles of [tex]\( \text{CO}_2 \)[/tex] is also 14.43 moles.
3. Calculate the mass of [tex]\( \text{CO} \)[/tex] required:
To find the mass, convert the moles of [tex]\( \text{CO} \)[/tex] back into grams using the molar mass of [tex]\( \text{CO} \)[/tex]:
[tex]\[
\text{Mass of } \text{CO} = \text{number of moles of } \text{CO} \times \text{molar mass of } \text{CO}
\][/tex]
[tex]\[
\text{Mass of } \text{CO} = 14.43 \text{ moles} \times 28.01 \text{ g/mol} \approx 404.14 \text{ g}
\][/tex]
### Conclusion
Therefore, 404.14 grams of CO would be required to generate 635 grams of [tex]\( \text{CO}_2 \)[/tex].
