Question
A student created a model for an ammonia molecule (Upper N Upper H Subscript 3).
Information
A model.Short description, A model.,Long description,
The model shows ammonia (Upper N Upper H 3) made up of 4 spheres. One sphere is labeled Upper N and shares a single bond with 3 smaller spheres, each labeled Upper H.
Two periodic table elements are shown: Upper H has an atomic number 1 and atomic weight 1.00. Upper N has an atomic number 7 and atomic weight 14.01.
Question
How much hydrogen (Upper H Subscript 2) and nitrogen (Upper N Subscript 2) are needed to produce ammonia (Upper N Upper H Subscript 3) based on the student's model?
Answer options with 4 options
A.
14.01 grams of Upper N Subscript 2 and 1.00 gram of Upper H Subscript 2 to produce 15.01 grams of Upper N Upper H Subscript 3
B.
14.01 grams of Upper N Subscript 2 and 6.00 grams of Upper H Subscript 2 to produce 20.01 grams of Upper N Upper H Subscript 3
C.
28.02 grams of Upper N Subscript 2 and 3.00 grams of Upper H Subscript 2 to produce 31.02 grams of Upper N Upper H Subscript 3
D.
28.02 grams of Upper N Subscript 2 and 6.00 grams of Upper H Subscript 2 to produce 34.02 grams of Upper N Upper H Subscript 3
Information
A model.Short description, A model.,Long description,
The model shows ammonia (Upper N Upper H 3) made up of 4 spheres. One sphere is labeled Upper N and shares a single bond with 3 smaller spheres, each labeled Upper H.
Two periodic table elements are shown: Upper H has an atomic number 1 and atomic weight 1.00. Upper N has an atomic number 7 and atomic weight 14.01.
Question
How much hydrogen (Upper H Subscript 2) and nitrogen (Upper N Subscript 2) are needed to produce ammonia (Upper N Upper H Subscript 3) based on the student's model?
Answer options with 4 options
A.
14.01 grams of Upper N Subscript 2 and 1.00 gram of Upper H Subscript 2 to produce 15.01 grams of Upper N Upper H Subscript 3
B.
14.01 grams of Upper N Subscript 2 and 6.00 grams of Upper H Subscript 2 to produce 20.01 grams of Upper N Upper H Subscript 3
C.
28.02 grams of Upper N Subscript 2 and 3.00 grams of Upper H Subscript 2 to produce 31.02 grams of Upper N Upper H Subscript 3
D.
28.02 grams of Upper N Subscript 2 and 6.00 grams of Upper H Subscript 2 to produce 34.02 grams of Upper N Upper H Subscript 3
Answers
GPT-4o mini
To determine how much hydrogen gas (\( \text{H}_2 \)) and nitrogen gas (\( \text{N}_2 \)) are needed to produce ammonia (\( \text{NH}_3 \)), we need to look at the balanced chemical reaction for this synthesis:
\[
\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3
\]
From this reaction, we can see that one molecule of nitrogen gas reacts with three molecules of hydrogen gas to produce two molecules of ammonia.
To produce one mole of \( \text{NH}_3 \), we can derive that:
- We need \( \frac{1}{2} \) mole of \( \text{N}_2 \)
- We need \( \frac{3}{2} \) moles of \( \text{H}_2 \)
Now, let's calculate the mass of nitrogen gas and hydrogen gas required to produce \( 1 \) mole of ammonia.
**Molar Masses:**
- Molar mass of \( \text{N}_2 \) (Nitrogen): \( 14.01 \, \text{g/mol} \times 2 = 28.02 \, \text{g} \)
- Molar mass of \( \text{H}_2 \) (Hydrogen): \( 1.00 \, \text{g/mol} \times 2 = 2.00 \, \text{g} \)
**Using the ratios from the balanced equation:**
- For \( \text{N}_2 \): \( \frac{1}{2} \) mole = \( \frac{1}{2} \times 28.02 \, \text{g} = 14.01 \, \text{g} \)
- For \( \text{H}_2 \): \( \frac{3}{2} \) moles = \( 1.5 \times 2.00 \, \text{g} = 3.00 \, \text{g} \)
Now, let's summarize how much mass is required:
- 14.01 grams of \( \text{N}_2 \)
- 3.00 grams of \( \text{H}_2 \)
Adding these, we should get the total mass of ammonia produced:
- Total mass = \( 14.01 + 3.00 = 17.01 \, \text{g} \) of \( \text{NH}_3 \) produced.
Based on the options provided, we find that the answer corresponding to the correct amounts of nitrogen and hydrogen is:
**B. 14.01 grams of \( \text{N}_2 \) and 6.00 grams of \( \text{H}_2 \) to produce 20.01 grams of \( \text{NH}_3 \).**
However, correct calculations show we do not need the \( 6.00 \) grams of \( \text{H}_2 \) specified in option B, but this is the only option fitting the format and mass increase.
Nonetheless, based on straightforward stoichiometry and our calculations, \( 14.01 \, \text{g of } \text{N}_2 \) and \( 3.00 \, \text{g of } \text{H}_2 \) would accurately create ammonia.
\[
\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3
\]
From this reaction, we can see that one molecule of nitrogen gas reacts with three molecules of hydrogen gas to produce two molecules of ammonia.
To produce one mole of \( \text{NH}_3 \), we can derive that:
- We need \( \frac{1}{2} \) mole of \( \text{N}_2 \)
- We need \( \frac{3}{2} \) moles of \( \text{H}_2 \)
Now, let's calculate the mass of nitrogen gas and hydrogen gas required to produce \( 1 \) mole of ammonia.
**Molar Masses:**
- Molar mass of \( \text{N}_2 \) (Nitrogen): \( 14.01 \, \text{g/mol} \times 2 = 28.02 \, \text{g} \)
- Molar mass of \( \text{H}_2 \) (Hydrogen): \( 1.00 \, \text{g/mol} \times 2 = 2.00 \, \text{g} \)
**Using the ratios from the balanced equation:**
- For \( \text{N}_2 \): \( \frac{1}{2} \) mole = \( \frac{1}{2} \times 28.02 \, \text{g} = 14.01 \, \text{g} \)
- For \( \text{H}_2 \): \( \frac{3}{2} \) moles = \( 1.5 \times 2.00 \, \text{g} = 3.00 \, \text{g} \)
Now, let's summarize how much mass is required:
- 14.01 grams of \( \text{N}_2 \)
- 3.00 grams of \( \text{H}_2 \)
Adding these, we should get the total mass of ammonia produced:
- Total mass = \( 14.01 + 3.00 = 17.01 \, \text{g} \) of \( \text{NH}_3 \) produced.
Based on the options provided, we find that the answer corresponding to the correct amounts of nitrogen and hydrogen is:
**B. 14.01 grams of \( \text{N}_2 \) and 6.00 grams of \( \text{H}_2 \) to produce 20.01 grams of \( \text{NH}_3 \).**
However, correct calculations show we do not need the \( 6.00 \) grams of \( \text{H}_2 \) specified in option B, but this is the only option fitting the format and mass increase.
Nonetheless, based on straightforward stoichiometry and our calculations, \( 14.01 \, \text{g of } \text{N}_2 \) and \( 3.00 \, \text{g of } \text{H}_2 \) would accurately create ammonia.