Question
1. Balance the following equation ___Fe(s) + ___O₂(g) ⟶ ___Fe₂O₃(s)
2. How many grams of iron (III) oxide are formed when 25 grams of iron reacts completely with oxygen?
2. How many grams of iron (III) oxide are formed when 25 grams of iron reacts completely with oxygen?
Answers
GPT-4o mini
### 1. Balancing the Equation
To balance the reaction equation for the formation of iron(III) oxide \( \text{Fe}_2\text{O}_3 \) from iron \( \text{Fe} \) and oxygen \( \text{O}_2 \), we start with the unbalanced equation:
\[ \_\_ \text{Fe}(s) + \_\_ \text{O}_2(g) \rightarrow \_\_ \text{Fe}_2\text{O}_3(s) \]
1. **Count the number of atoms for each element in the products:**
- In \( \text{Fe}_2\text{O}_3 \), there are 2 Fe atoms and 3 O atoms.
2. **Set the coefficients for products:**
- Put a coefficient of 1 in front of \( \text{Fe}_2\text{O}_3 \):
\[ \_\_ \text{Fe} + \_\_ \text{O}_2 \rightarrow 1 \text{Fe}_2\text{O}_3 \]
3. **Adjust the coefficient for Fe to balance iron:**
- Since there are 2 Fe in the product, change Fe to 4:
\[ 4 \text{Fe} + \_\_ \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3 \]
4. **Adjust the coefficient for O:**
- To balance the oxygen, since there are 6 O (3 O in \( \text{Fe}_2\text{O}_3 \) times 2) in the products, you need 3 \( \text{O}_2 \):
\[ 4 \text{Fe} + 3 \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3 \]
So the balanced equation is:
\[ 4 \text{Fe}(s) + 3 \text{O}_2(g) \rightarrow 2 \text{Fe}_2\text{O}_3(s) \]
### 2. Calculating Grams of Iron(III) Oxide Formed
To find how many grams of iron(III) oxide \( \text{Fe}_2\text{O}_3 \) are formed from 25 grams of iron \( \text{Fe} \):
1. **Find the molar mass of Fe:**
- Iron (Fe) has a molar mass of approximately 55.85 g/mol.
2. **Convert grams of Fe to moles:**
\[
\text{moles of Fe} = \frac{25 \text{ g}}{55.85 \text{ g/mol}} \approx 0.447 \text{ moles Fe}
\]
3. **Use the balanced equation to find moles of \( \text{Fe}_2\text{O}_3 \):**
- From the balanced equation \( 4 \text{Fe} + 3 \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3 \), we see that 4 moles of Fe produce 2 moles of \( \text{Fe}_2\text{O}_3 \).
Therefore, the ratio of Fe to \( \text{Fe}_2\text{O}_3 \) is \( \frac{2 \text{ moles } \text{Fe}_2\text{O}_3}{4 \text{ moles Fe}} = \frac{1}{2} \):
\[
\text{moles of } \text{Fe}_2\text{O}_3 = \frac{0.447 \text{ moles Fe}}{2} \approx 0.224 \text{ moles } \text{Fe}_2\text{O}_3
\]
4. **Find the molar mass of \( \text{Fe}_2\text{O}_3 \):**
- Molar mass \( \text{Fe}_2\text{O}_3 = 2(55.85 \text{ g/mol}) + 3(16.00 \text{ g/mol}) = 159.7 \text{ g/mol} \)
5. **Convert moles of \( \text{Fe}_2\text{O}_3 \) to grams:**
\[
\text{grams of } \text{Fe}_2\text{O}_3 = 0.224 \text{ moles} \times 159.7 \text{ g/mol} \approx 35.70 \text{ g}
\]
### Final Answer:
- The balanced equation is:
\[ 4 \text{Fe}(s) + 3 \text{O}_2(g) \rightarrow 2 \text{Fe}_2\text{O}_3(s) \]
- The amount of iron(III) oxide formed from 25 grams of iron is approximately **35.70 grams**.
To balance the reaction equation for the formation of iron(III) oxide \( \text{Fe}_2\text{O}_3 \) from iron \( \text{Fe} \) and oxygen \( \text{O}_2 \), we start with the unbalanced equation:
\[ \_\_ \text{Fe}(s) + \_\_ \text{O}_2(g) \rightarrow \_\_ \text{Fe}_2\text{O}_3(s) \]
1. **Count the number of atoms for each element in the products:**
- In \( \text{Fe}_2\text{O}_3 \), there are 2 Fe atoms and 3 O atoms.
2. **Set the coefficients for products:**
- Put a coefficient of 1 in front of \( \text{Fe}_2\text{O}_3 \):
\[ \_\_ \text{Fe} + \_\_ \text{O}_2 \rightarrow 1 \text{Fe}_2\text{O}_3 \]
3. **Adjust the coefficient for Fe to balance iron:**
- Since there are 2 Fe in the product, change Fe to 4:
\[ 4 \text{Fe} + \_\_ \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3 \]
4. **Adjust the coefficient for O:**
- To balance the oxygen, since there are 6 O (3 O in \( \text{Fe}_2\text{O}_3 \) times 2) in the products, you need 3 \( \text{O}_2 \):
\[ 4 \text{Fe} + 3 \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3 \]
So the balanced equation is:
\[ 4 \text{Fe}(s) + 3 \text{O}_2(g) \rightarrow 2 \text{Fe}_2\text{O}_3(s) \]
### 2. Calculating Grams of Iron(III) Oxide Formed
To find how many grams of iron(III) oxide \( \text{Fe}_2\text{O}_3 \) are formed from 25 grams of iron \( \text{Fe} \):
1. **Find the molar mass of Fe:**
- Iron (Fe) has a molar mass of approximately 55.85 g/mol.
2. **Convert grams of Fe to moles:**
\[
\text{moles of Fe} = \frac{25 \text{ g}}{55.85 \text{ g/mol}} \approx 0.447 \text{ moles Fe}
\]
3. **Use the balanced equation to find moles of \( \text{Fe}_2\text{O}_3 \):**
- From the balanced equation \( 4 \text{Fe} + 3 \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3 \), we see that 4 moles of Fe produce 2 moles of \( \text{Fe}_2\text{O}_3 \).
Therefore, the ratio of Fe to \( \text{Fe}_2\text{O}_3 \) is \( \frac{2 \text{ moles } \text{Fe}_2\text{O}_3}{4 \text{ moles Fe}} = \frac{1}{2} \):
\[
\text{moles of } \text{Fe}_2\text{O}_3 = \frac{0.447 \text{ moles Fe}}{2} \approx 0.224 \text{ moles } \text{Fe}_2\text{O}_3
\]
4. **Find the molar mass of \( \text{Fe}_2\text{O}_3 \):**
- Molar mass \( \text{Fe}_2\text{O}_3 = 2(55.85 \text{ g/mol}) + 3(16.00 \text{ g/mol}) = 159.7 \text{ g/mol} \)
5. **Convert moles of \( \text{Fe}_2\text{O}_3 \) to grams:**
\[
\text{grams of } \text{Fe}_2\text{O}_3 = 0.224 \text{ moles} \times 159.7 \text{ g/mol} \approx 35.70 \text{ g}
\]
### Final Answer:
- The balanced equation is:
\[ 4 \text{Fe}(s) + 3 \text{O}_2(g) \rightarrow 2 \text{Fe}_2\text{O}_3(s) \]
- The amount of iron(III) oxide formed from 25 grams of iron is approximately **35.70 grams**.