Question
The balanced chemical equation below represents the reaction of nitrogen (N₂) with hydrogen (H₂) to produce ammonia (NH₃):
N2 + 3H2 → 2NH3
If 10.0 grams of nitrogen reacts with hydrogen to yield 12.0 grams of ammonia, determine the mass of hydrogen used in the reaction?
(1 point)
Responses
10 g of H2
10 g of H 2
12 g of H2
12 g of H 2
3.0 g of H2
3.0 g of H 2
2.0 g of H2
N2 + 3H2 → 2NH3
If 10.0 grams of nitrogen reacts with hydrogen to yield 12.0 grams of ammonia, determine the mass of hydrogen used in the reaction?
(1 point)
Responses
10 g of H2
10 g of H 2
12 g of H2
12 g of H 2
3.0 g of H2
3.0 g of H 2
2.0 g of H2
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To determine the mass of hydrogen used in the reaction, we first need to find the molar masses of nitrogen (N₂), hydrogen (H₂), and ammonia (NH₃):
- The molar mass of N₂ (nitrogen) is approximately \( 28.0 \, \text{g/mol} \).
- The molar mass of H₂ (hydrogen) is approximately \( 2.0 \, \text{g/mol} \).
- The molar mass of NH₃ (ammonia) is approximately \( 17.0 \, \text{g/mol} \).
Next, we can calculate the moles of nitrogen and ammonia produced in the reaction:
1. Calculate the moles of N₂ used:
\[
\text{Moles of } N_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{10.0 \, \text{g}}{28.0 \, \text{g/mol}} \approx 0.357 \, \text{mol}
\]
2. According to the balanced equation \( N_2 + 3H_2 \rightarrow 2NH_3 \), 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃.
3. Calculate the moles of NH₃ produced:
\[
\text{Moles of } NH_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{12.0 \, \text{g}}{17.0 \, \text{g/mol}} \approx 0.706 \, \text{mol}
\]
4. From the stoichiometry of the reaction:
- From 1 mole of N₂, we produce 2 moles of NH₃.
- Thus, 0.357 moles of N₂ will produce:
\[
\text{Moles of } NH_3 = 0.357 \, \text{mol} \times 2 = 0.714 \, \text{mol}
\]
5. Since \( 0.706 \, \text{mol} \) of NH₃ is produced, this is consistent with our earlier calculation (indicating possibly some minor losses or differences in theoretical versus actual yields).
6. Next, we can determine the moles of H₂ needed:
- From the balanced equation, 1 mole of N₂ requires 3 moles of H₂.
- Therefore, \( 0.357 \) moles of N₂ would require:
\[
0.357 \, \text{mol} \times 3 = 1.071 \, \text{mol} \, H_2
\]
7. Calculate the mass of H₂ used:
\[
\text{Mass of } H_2 = \text{moles} \times \text{molar mass} = 1.071 \, \text{mol} \times 2.0 \, \text{g/mol} \approx 2.14 \, \text{g}
\]
Therefore, rounding the value, the mass of hydrogen used in the reaction is approximately:
**2.0 g of H₂**
- The molar mass of N₂ (nitrogen) is approximately \( 28.0 \, \text{g/mol} \).
- The molar mass of H₂ (hydrogen) is approximately \( 2.0 \, \text{g/mol} \).
- The molar mass of NH₃ (ammonia) is approximately \( 17.0 \, \text{g/mol} \).
Next, we can calculate the moles of nitrogen and ammonia produced in the reaction:
1. Calculate the moles of N₂ used:
\[
\text{Moles of } N_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{10.0 \, \text{g}}{28.0 \, \text{g/mol}} \approx 0.357 \, \text{mol}
\]
2. According to the balanced equation \( N_2 + 3H_2 \rightarrow 2NH_3 \), 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃.
3. Calculate the moles of NH₃ produced:
\[
\text{Moles of } NH_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{12.0 \, \text{g}}{17.0 \, \text{g/mol}} \approx 0.706 \, \text{mol}
\]
4. From the stoichiometry of the reaction:
- From 1 mole of N₂, we produce 2 moles of NH₃.
- Thus, 0.357 moles of N₂ will produce:
\[
\text{Moles of } NH_3 = 0.357 \, \text{mol} \times 2 = 0.714 \, \text{mol}
\]
5. Since \( 0.706 \, \text{mol} \) of NH₃ is produced, this is consistent with our earlier calculation (indicating possibly some minor losses or differences in theoretical versus actual yields).
6. Next, we can determine the moles of H₂ needed:
- From the balanced equation, 1 mole of N₂ requires 3 moles of H₂.
- Therefore, \( 0.357 \) moles of N₂ would require:
\[
0.357 \, \text{mol} \times 3 = 1.071 \, \text{mol} \, H_2
\]
7. Calculate the mass of H₂ used:
\[
\text{Mass of } H_2 = \text{moles} \times \text{molar mass} = 1.071 \, \text{mol} \times 2.0 \, \text{g/mol} \approx 2.14 \, \text{g}
\]
Therefore, rounding the value, the mass of hydrogen used in the reaction is approximately:
**2.0 g of H₂**
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