Question
N2(g)+3H2(g)=2NH3(g)
a.If you have 1.00*10^3 g of H2, which is the limiting reactant in the reaction?
b. How many grams of ammonia can be produced from the amount of limiting reactant available?
c. Calculate the mass of excess reactant that remains after the reaction is complete
a.If you have 1.00*10^3 g of H2, which is the limiting reactant in the reaction?
b. How many grams of ammonia can be produced from the amount of limiting reactant available?
c. Calculate the mass of excess reactant that remains after the reaction is complete
Answers
Answered by
GPT-4o mini
Let's solve the problem step-by-step.
**Given Reaction:**
\[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \]
**Molar Mass Calculation:**
- Molar mass of \( \text{H}_2 \) = 2.02 g/mol
- Molar mass of \( \text{N}_2 \) = 28.02 g/mol
- Molar mass of \( \text{NH}_3 \) = 17.03 g/mol
**Part A: Limiting Reactant Calculation**
1. Calculate the number of moles of \( \text{H}_2 \):
\[
\text{Moles of } \text{H}_2 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{1.00 \times 10^3 \text{ g}}{2.02 \text{ g/mol}} \approx 495.05 \text{ mol}
\]
2. According to the stoichiometry of the reaction:
- 3 moles of \( \text{H}_2 \) are required for every 1 mole of \( \text{N}_2 \).
- Thus, we need to determine how much \( \text{N}_2 \) is needed for the available \( \text{H}_2 \):
\[
\text{Moles of } \text{N}_2 = \frac{495.05 \text{ mol } \text{H}_2}{3} \approx 165.02 \text{ mol } \text{N}_2
\]
3. If we do not have a specified amount of \( \text{N}_2 \), we cannot definitively identify the limiting reactant without knowing the amount available. Assuming we have excess \( \text{N}_2 \) sufficient to react with all the \( \text{H}_2 \), then \( \text{H}_2 \) is the limiting reactant.
**Part B: Grams of Ammonia Produced**
1. From the balanced equation, 3 moles of \( \text{H}_2 \) produce 2 moles of \( \text{NH}_3 \):
\[
\text{Moles of } \text{NH}_3 = \frac{2}{3} \times \text{Moles of } \text{H}_2 = \frac{2}{3} \times 495.05 \approx 330.03 \text{ mol } \text{NH}_3
\]
2. Now, convert moles of \( \text{NH}_3 \) to grams:
\[
\text{Mass of } \text{NH}_3 = \text{Moles} \times \text{Molar Mass} = 330.03 \text{ mol} \times 17.03 \text{ g/mol} \approx 5626.91 \text{ g}
\]
**Part C: Mass of Excess Reactant Remaining**
1. Since \( \text{H}_2 \) is the limiting reactant, we need to find out how much \( \text{N}_2 \) is used.
\[
\text{Moles of } \text{N}_2 \text{ used} = \frac{1}{3} \times \text{Moles of } \text{H}_2 = \frac{1}{3} \times 495.05 \approx 165.02 \text{ mol } \text{N}_2
\]
2. Convert moles of \( \text{N}_2 \) used to grams:
\[
\text{Mass of } \text{N}_2 \text{ used} = 165.02 \text{ mol} \times 28.02 \text{ g/mol} \approx 4626.60 \text{ g}
\]
3. If we are given a specific amount of \( \text{N}_2 \) at the start, we can calculate the remaining amount. Let's assume we started with enough \( \text{N}_2 \) that it was in excess. Without knowing the initial amount of \( \text{N}_2 \), we can't determine the remaining mass of \( \text{N}_2 \).
In summary, if you have lots of \( \text{N}_2 \):
- Limiting Reactant: \( \text{H}_2 \)
- Ammonia produced: \( \approx 5626.91 \text{ g} \)
- Remaining excess reactant \( \text{N}_2 \) cannot be determined without additional data on the initial amount of \( \text{N}_2 \).
**Given Reaction:**
\[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \]
**Molar Mass Calculation:**
- Molar mass of \( \text{H}_2 \) = 2.02 g/mol
- Molar mass of \( \text{N}_2 \) = 28.02 g/mol
- Molar mass of \( \text{NH}_3 \) = 17.03 g/mol
**Part A: Limiting Reactant Calculation**
1. Calculate the number of moles of \( \text{H}_2 \):
\[
\text{Moles of } \text{H}_2 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{1.00 \times 10^3 \text{ g}}{2.02 \text{ g/mol}} \approx 495.05 \text{ mol}
\]
2. According to the stoichiometry of the reaction:
- 3 moles of \( \text{H}_2 \) are required for every 1 mole of \( \text{N}_2 \).
- Thus, we need to determine how much \( \text{N}_2 \) is needed for the available \( \text{H}_2 \):
\[
\text{Moles of } \text{N}_2 = \frac{495.05 \text{ mol } \text{H}_2}{3} \approx 165.02 \text{ mol } \text{N}_2
\]
3. If we do not have a specified amount of \( \text{N}_2 \), we cannot definitively identify the limiting reactant without knowing the amount available. Assuming we have excess \( \text{N}_2 \) sufficient to react with all the \( \text{H}_2 \), then \( \text{H}_2 \) is the limiting reactant.
**Part B: Grams of Ammonia Produced**
1. From the balanced equation, 3 moles of \( \text{H}_2 \) produce 2 moles of \( \text{NH}_3 \):
\[
\text{Moles of } \text{NH}_3 = \frac{2}{3} \times \text{Moles of } \text{H}_2 = \frac{2}{3} \times 495.05 \approx 330.03 \text{ mol } \text{NH}_3
\]
2. Now, convert moles of \( \text{NH}_3 \) to grams:
\[
\text{Mass of } \text{NH}_3 = \text{Moles} \times \text{Molar Mass} = 330.03 \text{ mol} \times 17.03 \text{ g/mol} \approx 5626.91 \text{ g}
\]
**Part C: Mass of Excess Reactant Remaining**
1. Since \( \text{H}_2 \) is the limiting reactant, we need to find out how much \( \text{N}_2 \) is used.
\[
\text{Moles of } \text{N}_2 \text{ used} = \frac{1}{3} \times \text{Moles of } \text{H}_2 = \frac{1}{3} \times 495.05 \approx 165.02 \text{ mol } \text{N}_2
\]
2. Convert moles of \( \text{N}_2 \) used to grams:
\[
\text{Mass of } \text{N}_2 \text{ used} = 165.02 \text{ mol} \times 28.02 \text{ g/mol} \approx 4626.60 \text{ g}
\]
3. If we are given a specific amount of \( \text{N}_2 \) at the start, we can calculate the remaining amount. Let's assume we started with enough \( \text{N}_2 \) that it was in excess. Without knowing the initial amount of \( \text{N}_2 \), we can't determine the remaining mass of \( \text{N}_2 \).
In summary, if you have lots of \( \text{N}_2 \):
- Limiting Reactant: \( \text{H}_2 \)
- Ammonia produced: \( \approx 5626.91 \text{ g} \)
- Remaining excess reactant \( \text{N}_2 \) cannot be determined without additional data on the initial amount of \( \text{N}_2 \).