Question
Silicon nitride is a very hard, high-temperature-resistant ceramic used as a component of turbine blades in jet engines. It is prepared according to the following equation: 3Si(s) + 2N2
(g) ⟶ Si3N4(s)
Which is the limiting reactant when 2.00 g of Si and 1.50 g of N2 react?
(g) ⟶ Si3N4(s)
Which is the limiting reactant when 2.00 g of Si and 1.50 g of N2 react?
Answers
GPT-4o mini
To identify the limiting reactant in the reaction \(3 \text{Si}(s) + 2 \text{N}_2(g) \rightarrow \text{Si}_3\text{N}_4(s)\), we'll first need to convert the masses of silicon (Si) and nitrogen (N\(_2\)) gas into moles.
### Step 1: Calculate the moles of each reactant.
1. **Calculate moles of Si:**
- Molar mass of Si = 28.09 g/mol
\[
\text{moles of Si} = \frac{\text{mass}}{\text{molar mass}} = \frac{2.00 \, \text{g}}{28.09 \, \text{g/mol}} \approx 0.0712 \, \text{mol}
\]
2. **Calculate moles of N\(_2\):**
- Molar mass of N\(_2\) = 28.02 g/mol
\[
\text{moles of } N_2 = \frac{1.50 \, \text{g}}{28.02 \, \text{g/mol}} \approx 0.0535 \, \text{mol}
\]
### Step 2: Determine the stoichiometric requirements of the reaction.
From the balanced reaction \(3 \text{Si} + 2 \text{N}_2 \rightarrow \text{Si}_3\text{N}_4\), we see that:
- 3 moles of Si react with 2 moles of N\(_2\).
### Step 3: Calculate the required amounts for each reactant.
We have the following mole ratios from the reaction:
- For every 3 moles of Si, we need 2 moles of N\(_2\).
- This can be rearranged to find out how much of one reactant is needed to react with the other.
1. **For Si:**
- From the moles of Si, calculate the moles of N\(_2\) required:
\[
\text{moles of } N_2 \text{ required} = \frac{2}{3} \times \text{moles of Si} = \frac{2}{3} \times 0.0712 \approx 0.0475 \, \text{mol}
\]
2. **For N\(_2\):**
- From the moles of N\(_2\), calculate the moles of Si that can react:
\[
\text{moles of Si required} = \frac{3}{2} \times \text{moles of } N_2 = \frac{3}{2} \times 0.0535 \approx 0.0803 \, \text{mol}
\]
### Step 4: Compare the available amounts to the required amounts.
- We have approximately 0.0712 moles of Si, but we only need about 0.0475 moles of N\(_2\) based on the available Si, which is sufficient since we have 0.0535 moles of N\(_2\) available.
- However, we require approximately 0.0803 moles of Si to react with all of the N\(_2\), but we only have 0.0712 moles available.
### Conclusion: Identify the limiting reactant.
Since we do not have enough Si to react with all of the N\(_2\), silicon (Si) is the limiting reactant. Thus, **the limiting reactant is silicon (Si)**.
### Step 1: Calculate the moles of each reactant.
1. **Calculate moles of Si:**
- Molar mass of Si = 28.09 g/mol
\[
\text{moles of Si} = \frac{\text{mass}}{\text{molar mass}} = \frac{2.00 \, \text{g}}{28.09 \, \text{g/mol}} \approx 0.0712 \, \text{mol}
\]
2. **Calculate moles of N\(_2\):**
- Molar mass of N\(_2\) = 28.02 g/mol
\[
\text{moles of } N_2 = \frac{1.50 \, \text{g}}{28.02 \, \text{g/mol}} \approx 0.0535 \, \text{mol}
\]
### Step 2: Determine the stoichiometric requirements of the reaction.
From the balanced reaction \(3 \text{Si} + 2 \text{N}_2 \rightarrow \text{Si}_3\text{N}_4\), we see that:
- 3 moles of Si react with 2 moles of N\(_2\).
### Step 3: Calculate the required amounts for each reactant.
We have the following mole ratios from the reaction:
- For every 3 moles of Si, we need 2 moles of N\(_2\).
- This can be rearranged to find out how much of one reactant is needed to react with the other.
1. **For Si:**
- From the moles of Si, calculate the moles of N\(_2\) required:
\[
\text{moles of } N_2 \text{ required} = \frac{2}{3} \times \text{moles of Si} = \frac{2}{3} \times 0.0712 \approx 0.0475 \, \text{mol}
\]
2. **For N\(_2\):**
- From the moles of N\(_2\), calculate the moles of Si that can react:
\[
\text{moles of Si required} = \frac{3}{2} \times \text{moles of } N_2 = \frac{3}{2} \times 0.0535 \approx 0.0803 \, \text{mol}
\]
### Step 4: Compare the available amounts to the required amounts.
- We have approximately 0.0712 moles of Si, but we only need about 0.0475 moles of N\(_2\) based on the available Si, which is sufficient since we have 0.0535 moles of N\(_2\) available.
- However, we require approximately 0.0803 moles of Si to react with all of the N\(_2\), but we only have 0.0712 moles available.
### Conclusion: Identify the limiting reactant.
Since we do not have enough Si to react with all of the N\(_2\), silicon (Si) is the limiting reactant. Thus, **the limiting reactant is silicon (Si)**.