How Many Grams Of N H 3 NH_3 N H 3 ​ Can Be Produced From The Reaction Of 28 G Of N 2 N_2 N 2 ​ And 25 G Of H 2 H_2 H 2 ​ ?Reaction: N 2 + 3 H 2 → 2 N H 3 N_2 + 3H_2 \rightarrow 2NH_3 N 2 ​ + 3 H 2 ​ → 2 N H 3 ​

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Understanding the Chemical Reaction

The given chemical reaction is $N_2 + 3H_2 \rightarrow 2NH_3$. This reaction involves the combination of nitrogen gas ($N_2$) and hydrogen gas ($H_2$) to produce ammonia ($NH_3$). The balanced equation indicates that 1 mole of $N_2$ reacts with 3 moles of $H_2$ to produce 2 moles of $NH_3$.

Calculating the Number of Moles of $N_2$ and $H_2$

To determine the number of grams of $NH_3$ that can be produced, we need to calculate the number of moles of $N_2$ and $H_2$ present in the given quantities.

The molar mass of $N_2$ is 28 g/mol, and the molar mass of $H_2$ is 2 g/mol. We can calculate the number of moles of $N_2$ and $H_2$ using the following formulas:

• Number of moles of $N_2$ = mass of $N_2$ / molar mass of $N_2$
• Number of moles of $H_2$ = mass of $H_2$ / molar mass of $H_2$

Plugging in the values, we get:

• Number of moles of $N_2$ = 28 g / 28 g/mol = 1 mol
• Number of moles of $H_2$ = 25 g / 2 g/mol = 12.5 mol

Determining the Limiting Reactant

Since the reaction requires a 1:3 ratio of $N_2$ to $H_2$, we need to determine which reactant is the limiting reactant. The limiting reactant is the reactant that will be completely consumed first, and it will determine the maximum amount of product that can be formed.

In this case, we have 1 mol of $N_2$ and 12.5 mol of $H_2$. Since the reaction requires 3 moles of $H_2$ for every 1 mole of $N_2$, we can see that the 1 mol of $N_2$ will be completely consumed by 3.75 mol of $H_2$ (1 mol x 3 = 3 mol, but we have 12.5 mol of $H_2$). Therefore, the $N_2$ is the limiting reactant.

Calculating the Number of Moles of $NH_3$ Produced

Since the $N_2$ is the limiting reactant, we can calculate the number of moles of $NH_3$ produced using the following formula:

• Number of moles of $NH_3$ = 2 x number of moles of $N_2$

Plugging in the value, we get:

• Number of moles of $NH_3$ = 2 x 1 mol = 2 mol

Calculating the Mass of $NH_3$ Produced

The molar mass of $NH_3$ is 17 g/mol. We can calculate the mass of $NH_3$ produced using the following formula:

• Mass of $NH_3$ = number of moles of $NH_3$ x molar mass of $NH_3$

Plugging in the values, we get:

• Mass of $NH_3$ = 2 mol x 17 g/mol = 34 g

Therefore, the maximum amount of $NH_3$ that can be produced from the reaction of 28 g of $N_2$ and 25 g of $H_2$ is 34 g.

Conclusion

In conclusion, the reaction of 28 g of $N_2$ and 25 g of $H_2$ can produce a maximum of 34 g of $NH_3$. This calculation is based on the balanced chemical equation and the molar masses of the reactants and products. The limiting reactant in this case is the $N_2$, and the number of moles of $NH_3$ produced is determined by the number of moles of $N_2$ present.

Limitations of the Calculation

It's worth noting that this calculation assumes that the reaction is carried out in a perfectly ideal environment, with no losses or inefficiencies. In reality, there may be losses due to factors such as heat transfer, mass transfer, and reaction kinetics. Therefore, the actual yield of $NH_3$ may be lower than the calculated value.

Future Directions

This calculation can be used as a starting point for further analysis and optimization of the reaction. For example, we could investigate the effects of different reaction conditions, such as temperature and pressure, on the yield of $NH_3$. We could also explore the use of catalysts or other additives to improve the reaction efficiency.

References

• [1] "Chemical Reaction Engineering" by Octave Levenspiel
• [2] "Chemical Thermodynamics" by John W. Moore
• [3] "Physical Chemistry" by Peter Atkins and Julio de Paula

Note: The references provided are for illustrative purposes only and are not intended to be a comprehensive list of sources.

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Q: What is the balanced chemical equation for the reaction of $N_2$ and $H_2$ to produce $NH_3$?

A: The balanced chemical equation for the reaction is $N_2 + 3H_2 \rightarrow 2NH_3$.

Q: What is the molar mass of $N_2$ and $H_2$?

A: The molar mass of $N_2$ is 28 g/mol, and the molar mass of $H_2$ is 2 g/mol.

Q: How do you calculate the number of moles of $N_2$ and $H_2$ present in a given quantity?

A: To calculate the number of moles of $N_2$ and $H_2$, you can use the following formulas:

• Number of moles of $N_2$ = mass of $N_2$ / molar mass of $N_2$
• Number of moles of $H_2$ = mass of $H_2$ / molar mass of $H_2$

Q: What is the limiting reactant in the reaction of $N_2$ and $H_2$ to produce $NH_3$?

A: In this case, the $N_2$ is the limiting reactant.

Q: How do you calculate the number of moles of $NH_3$ produced in the reaction?

A: To calculate the number of moles of $NH_3$ produced, you can use the following formula:

• Number of moles of $NH_3$ = 2 x number of moles of $N_2$

Q: What is the molar mass of $NH_3$?

A: The molar mass of $NH_3$ is 17 g/mol.

Q: How do you calculate the mass of $NH_3$ produced in the reaction?

A: To calculate the mass of $NH_3$ produced, you can use the following formula:

• Mass of $NH_3$ = number of moles of $NH_3$ x molar mass of $NH_3$

Q: What is the maximum amount of $NH_3$ that can be produced from the reaction of 28 g of $N_2$ and 25 g of $H_2$?

A: The maximum amount of $NH_3$ that can be produced is 34 g.

Q: What are some limitations of the calculation?

A: The calculation assumes that the reaction is carried out in a perfectly ideal environment, with no losses or inefficiencies. In reality, there may be losses due to factors such as heat transfer, mass transfer, and reaction kinetics.

Q: What are some future directions for further analysis and optimization of the reaction?

A: Some possible future directions include investigating the effects of different reaction conditions, such as temperature and pressure, on the yield of $NH_3$, and exploring the use of catalysts or other additives to improve the reaction efficiency.

Q: What are some references that can be used for further study?

A: Some references that can be used for further study include "Chemical Reaction Engineering" by Octave Levenspiel, "Chemical Thermodynamics" by John W. Moore, and "Physical Chemistry" by Peter Atkins and Julio de Paula.

Q: What is the significance of the reaction of $N_2$ and $H_2$ to produce $NH_3$?

A: The reaction of $N_2$ and $H_2$ to produce $NH_3$ is an important industrial process that is used to produce ammonia, which is a key component in the production of fertilizers, explosives, and other chemicals.

Q: What are some potential applications of the reaction of $N_2$ and $H_2$ to produce $NH_3$?

A: Some potential applications of the reaction include the production of fertilizers, explosives, and other chemicals, as well as the use of ammonia as a fuel source in power generation and transportation.

Q: What are some potential challenges associated with the reaction of $N_2$ and $H_2$ to produce $NH_3$?

A: Some potential challenges associated with the reaction include the need for high temperatures and pressures, the potential for catalyst poisoning, and the need for careful control of reaction conditions to optimize yield and selectivity.