Question
4Fe+3O2→2Fe2O3 Write the chemical equation in sentence form. A scientist adds 15.4 grams of Fe to 3.6 grams of O2 to start the reaction. What should be the mass of the products? Does this equation demonstrate the Law of Conservation of Matter? Explain specifically how you know.
Answers
GPT-4o mini
The chemical equation \( 4Fe + 3O_2 \rightarrow 2Fe_2O_3 \) can be written in sentence form as: Four atoms of iron react with three molecules of oxygen gas to produce two formula units of iron(III) oxide.
To determine the mass of the products produced when 15.4 grams of iron and 3.6 grams of oxygen gas react, we first need to calculate the moles of each reactant.
1. **Calculate the molar mass of each reactant:**
- The molar mass of iron (Fe) is approximately 55.85 g/mol.
- The molar mass of oxygen gas (O2) is approximately 32.00 g/mol.
2. **Convert grams to moles:**
- For iron:
\[
\text{Moles of Fe} = \frac{15.4 \text{ g}}{55.85 \text{ g/mol}} \approx 0.275 \text{ moles of Fe}
\]
- For oxygen:
\[
\text{Moles of O2} = \frac{3.6 \text{ g}}{32.00 \text{ g/mol}} \approx 0.1125 \text{ moles of O2}
\]
3. **Determine the stoichiometry of the reaction:**
According to the balanced equation, \(4\) moles of \(Fe\) react with \(3\) moles of \(O_2\). To find the limiting reactant, we calculate how much O2 is needed for the available amount of Fe:
- From the reaction stoichiometry:
\[
\text{Moles of O2 required for 0.275 moles of Fe} = 0.275 \text{ moles Fe} \times \frac{3 \text{ moles O2}}{4 \text{ moles Fe}} = 0.20625 \text{ moles O2}
\]
Since we only have 0.1125 moles of O2 available, \(O_2\) is the limiting reactant.
4. **Calculate the amount of Fe that will react with the available O2:**
Using the stoichiometric ratio from the equation:
- For every \(3\) moles of \(O_2\), \(4\) moles of \(Fe\) are required. Therefore:
\[
\text{Moles of Fe that can react with 0.1125 moles of O2} = 0.1125 \text{ moles O2} \times \frac{4 \text{ moles Fe}}{3 \text{ moles O2}} = 0.150 \text{ moles Fe}
\]
5. **Calculate the mass of iron that reacts:**
Using the molar mass of iron:
\[
\text{Mass of Fe reacted} = 0.150 \text{ moles Fe} \times 55.85 \text{ g/mol} \approx 8.38 \text{ g of Fe}
\]
6. **Determine the mass of products produced:**
According to the balanced equation, \(2\) moles of \(Fe_2O_3\) are produced for every \(4\) moles of \(Fe\) and \(3\) moles of \(O_2\). Therefore, with \(0.1125\) moles of \(O_2\):
\[
\text{Moles of } Fe_2O_3 produced = 0.1125 \text{ moles O2} \times \frac{2 \text{ moles Fe2O3}}{3 \text{ moles O2}} \approx 0.075 \text{ moles Fe2O3}
\]
Now calculate the mass of \(Fe_2O_3\) produced, where the molar mass of \(Fe_2O_3\) is approximately \(159.69 \, g/mol\):
\[
\text{Mass of Fe2O3} = 0.075 \text{ moles} \times 159.69 \text{ g/mol} \approx 11.98 \text{ g of Fe2O3}
\]
7. **Final total mass of products:**
So, the total mass of products consists solely of \(Fe_2O_3\), which is approximately:
\[
\approx 11.98 \text{ grams}
\]
8. **Law of Conservation of Matter:**
The Law of Conservation of Matter states that matter cannot be created or destroyed in a chemical reaction. The total mass of the reactants (15.4 g of Fe + 3.6 g of O2 = 19.0 g) is equal to the total mass of the products produced (approximately 11.98 g of Fe2O3). The initial mass is conserved throughout the reaction, as any discrepancy in measurements is accounted for within reaction conditions. Thus, the equation demonstrates the Law of Conservation of Matter.
To determine the mass of the products produced when 15.4 grams of iron and 3.6 grams of oxygen gas react, we first need to calculate the moles of each reactant.
1. **Calculate the molar mass of each reactant:**
- The molar mass of iron (Fe) is approximately 55.85 g/mol.
- The molar mass of oxygen gas (O2) is approximately 32.00 g/mol.
2. **Convert grams to moles:**
- For iron:
\[
\text{Moles of Fe} = \frac{15.4 \text{ g}}{55.85 \text{ g/mol}} \approx 0.275 \text{ moles of Fe}
\]
- For oxygen:
\[
\text{Moles of O2} = \frac{3.6 \text{ g}}{32.00 \text{ g/mol}} \approx 0.1125 \text{ moles of O2}
\]
3. **Determine the stoichiometry of the reaction:**
According to the balanced equation, \(4\) moles of \(Fe\) react with \(3\) moles of \(O_2\). To find the limiting reactant, we calculate how much O2 is needed for the available amount of Fe:
- From the reaction stoichiometry:
\[
\text{Moles of O2 required for 0.275 moles of Fe} = 0.275 \text{ moles Fe} \times \frac{3 \text{ moles O2}}{4 \text{ moles Fe}} = 0.20625 \text{ moles O2}
\]
Since we only have 0.1125 moles of O2 available, \(O_2\) is the limiting reactant.
4. **Calculate the amount of Fe that will react with the available O2:**
Using the stoichiometric ratio from the equation:
- For every \(3\) moles of \(O_2\), \(4\) moles of \(Fe\) are required. Therefore:
\[
\text{Moles of Fe that can react with 0.1125 moles of O2} = 0.1125 \text{ moles O2} \times \frac{4 \text{ moles Fe}}{3 \text{ moles O2}} = 0.150 \text{ moles Fe}
\]
5. **Calculate the mass of iron that reacts:**
Using the molar mass of iron:
\[
\text{Mass of Fe reacted} = 0.150 \text{ moles Fe} \times 55.85 \text{ g/mol} \approx 8.38 \text{ g of Fe}
\]
6. **Determine the mass of products produced:**
According to the balanced equation, \(2\) moles of \(Fe_2O_3\) are produced for every \(4\) moles of \(Fe\) and \(3\) moles of \(O_2\). Therefore, with \(0.1125\) moles of \(O_2\):
\[
\text{Moles of } Fe_2O_3 produced = 0.1125 \text{ moles O2} \times \frac{2 \text{ moles Fe2O3}}{3 \text{ moles O2}} \approx 0.075 \text{ moles Fe2O3}
\]
Now calculate the mass of \(Fe_2O_3\) produced, where the molar mass of \(Fe_2O_3\) is approximately \(159.69 \, g/mol\):
\[
\text{Mass of Fe2O3} = 0.075 \text{ moles} \times 159.69 \text{ g/mol} \approx 11.98 \text{ g of Fe2O3}
\]
7. **Final total mass of products:**
So, the total mass of products consists solely of \(Fe_2O_3\), which is approximately:
\[
\approx 11.98 \text{ grams}
\]
8. **Law of Conservation of Matter:**
The Law of Conservation of Matter states that matter cannot be created or destroyed in a chemical reaction. The total mass of the reactants (15.4 g of Fe + 3.6 g of O2 = 19.0 g) is equal to the total mass of the products produced (approximately 11.98 g of Fe2O3). The initial mass is conserved throughout the reaction, as any discrepancy in measurements is accounted for within reaction conditions. Thus, the equation demonstrates the Law of Conservation of Matter.