How Many Grams Of $NH_3$ Are Produced By The Reaction Of 5.40 Grams Of $H_2$?

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Introduction

In this article, we will explore the production of ammonia ($NH_3$) through the reaction of hydrogen gas ($H_2$) with nitrogen gas ($N_2$). The reaction is as follows:

$$N_2 + 3H_2 \rightarrow 2NH_3$$

We will calculate the number of grams of $NH_3$ produced by the reaction of 5.40 grams of $H_2$.

Theoretical Background

The reaction of nitrogen gas with hydrogen gas to produce ammonia is a well-known reaction in chemistry. The balanced equation for the reaction is given above. The reaction is highly exothermic and is used in the production of ammonia on a large scale.

Calculations

To calculate the number of grams of $NH_3$ produced, we need to know the molar masses of $H_2$ and $NH_3$. The molar masses are as follows:

• $H_2$: 2.02 g/mol
• $NH_3$: 17.03 g/mol

We are given 5.40 grams of $H_2$. We need to convert this to moles of $H_2$.

$$moles\ of\ H_2 = \frac{mass\ of\ H_2}{molar\ mass\ of\ H_2}$$

$$moles\ of\ H_2 = \frac{5.40\ g}{2.02\ g/mol}$$

$$moles\ of\ H_2 = 2.67\ mol$$

From the balanced equation, we can see that 3 moles of $H_2$ produce 2 moles of $NH_3$. Therefore, the number of moles of $NH_3$ produced is:

$$moles\ of\ NH_3 = \frac{2}{3} \times moles\ of\ H_2$$

$$moles\ of\ NH_3 = \frac{2}{3} \times 2.67\ mol$$

$$moles\ of\ NH_3 = 1.78\ mol$$

Now, we can calculate the mass of $NH_3$ produced by multiplying the number of moles of $NH_3$ by the molar mass of $NH_3$.

$$mass\ of\ NH_3 = moles\ of\ NH_3 \times molar\ mass\ of\ NH_3$$

$$mass\ of\ NH_3 = 1.78\ mol \times 17.03\ g/mol$$

$$mass\ of\ NH_3 = 30.3\ g$$

Conclusion

In this article, we calculated the number of grams of $NH_3$ produced by the reaction of 5.40 grams of $H_2$. We used the balanced equation for the reaction and the molar masses of $H_2$ and $NH_3$ to calculate the number of moles of $NH_3$ produced and then multiplied this by the molar mass of $NH_3$ to get the mass of $NH_3$ produced.

Limitations

This calculation assumes that the reaction goes to completion and that there are no side reactions. In reality, the reaction may not go to completion and there may be side reactions that affect the yield of $NH_3$.

Future Work

This calculation can be used as a starting point for further calculations. For example, we could calculate the energy released during the reaction or the pressure and temperature of the reaction.

References

• [1] "Chemical Equilibrium" by John W. Moore, Christian E. A. Kierstead, and John O. Johnston
• [2] "General Chemistry" by Linus Pauling

Appendix

The following is a list of the equations used in this article:

• $N_2 + 3H_2 \rightarrow 2NH_3$
• $moles\ of\ H_2 = \frac{mass\ of\ H_2}{molar\ mass\ of\ H_2}$
• $moles\ of\ NH_3 = \frac{2}{3} \times moles\ of\ H_2$
• $mass\ of\ NH_3 = moles\ of\ NH_3 \times molar\ mass\ of\ NH_3$
  Q&A: How Many Grams of $NH_3$ are Produced by the Reaction of 5.40 Grams of $H_2$? ====================================================================

Frequently Asked Questions

Q: What is the reaction of nitrogen gas with hydrogen gas to produce ammonia?

A: The reaction of nitrogen gas with hydrogen gas to produce ammonia is a well-known reaction in chemistry. The balanced equation for the reaction is:

$$N_2 + 3H_2 \rightarrow 2NH_3$$

Q: What is the molar mass of $H_2$ and $NH_3$?

A: The molar masses of $H_2$ and $NH_3$ are as follows:

• $H_2$: 2.02 g/mol
• $NH_3$: 17.03 g/mol

Q: How many moles of $H_2$ are in 5.40 grams of $H_2$?

A: To calculate the number of moles of $H_2$, we use the following equation:

$$moles\ of\ H_2 = \frac{mass\ of\ H_2}{molar\ mass\ of\ H_2}$$

$$moles\ of\ H_2 = \frac{5.40\ g}{2.02\ g/mol}$$

$$moles\ of\ H_2 = 2.67\ mol$$

Q: How many moles of $NH_3$ are produced by the reaction of 5.40 grams of $H_2$?

A: From the balanced equation, we can see that 3 moles of $H_2$ produce 2 moles of $NH_3$. Therefore, the number of moles of $NH_3$ produced is:

$$moles\ of\ NH_3 = \frac{2}{3} \times moles\ of\ H_2$$

$$moles\ of\ NH_3 = \frac{2}{3} \times 2.67\ mol$$

$$moles\ of\ NH_3 = 1.78\ mol$$

Q: How many grams of $NH_3$ are produced by the reaction of 5.40 grams of $H_2$?

A: To calculate the mass of $NH_3$ produced, we multiply the number of moles of $NH_3$ by the molar mass of $NH_3$.

$$mass\ of\ NH_3 = moles\ of\ NH_3 \times molar\ mass\ of\ NH_3$$

$$mass\ of\ NH_3 = 1.78\ mol \times 17.03\ g/mol$$

$$mass\ of\ NH_3 = 30.3\ g$$

Q: What are the limitations of this calculation?

A: This calculation assumes that the reaction goes to completion and that there are no side reactions. In reality, the reaction may not go to completion and there may be side reactions that affect the yield of $NH_3$.

Q: What are some potential future directions for this calculation?

A: This calculation can be used as a starting point for further calculations. For example, we could calculate the energy released during the reaction or the pressure and temperature of the reaction.

Additional Resources

• [1] "Chemical Equilibrium" by John W. Moore, Christian E. A. Kierstead, and John O. Johnston
• [2] "General Chemistry" by Linus Pauling

Glossary

• Molar mass: The mass of one mole of a substance.
• Mole: A unit of measurement for the amount of a substance.
• Reaction: A process in which one or more substances are converted into one or more other substances.

FAQs

• Q: What is the reaction of nitrogen gas with hydrogen gas to produce ammonia? A: The reaction of nitrogen gas with hydrogen gas to produce ammonia is a well-known reaction in chemistry. The balanced equation for the reaction is:

$$N_2 + 3H_2 \rightarrow 2NH_3$$

• Q: What is the molar mass of $H_2$ and $NH_3$? A: The molar masses of $H_2$ and $NH_3$ are as follows:

• $H_2$: 2.02 g/mol

• $NH_3$: 17.03 g/mol

• Q: How many moles of $H_2$ are in 5.40 grams of $H_2$? A: To calculate the number of moles of $H_2$, we use the following equation:

$$moles\ of\ H_2 = \frac{mass\ of\ H_2}{molar\ mass\ of\ H_2}$$

$$moles\ of\ H_2 = \frac{5.40\ g}{2.02\ g/mol}$$

$$moles\ of\ H_2 = 2.67\ mol$$

• Q: How many moles of $NH_3$ are produced by the reaction of 5.40 grams of $H_2$? A: From the balanced equation, we can see that 3 moles of $H_2$ produce 2 moles of $NH_3$. Therefore, the number of moles of $NH_3$ produced is:

$$moles\ of\ NH_3 = \frac{2}{3} \times moles\ of\ H_2$$

$$moles\ of\ NH_3 = \frac{2}{3} \times 2.67\ mol$$

$$moles\ of\ NH_3 = 1.78\ mol$$

• Q: How many grams of $NH_3$ are produced by the reaction of 5.40 grams of $H_2$? A: To calculate the mass of $NH_3$ produced, we multiply the number of moles of $NH_3$ by the molar mass of $NH_3$.

$$mass\ of\ NH_3 = moles\ of\ NH_3 \times molar\ mass\ of\ NH_3$$

$$mass\ of\ NH_3 = 1.78\ mol \times 17.03\ g/mol$$

$$mass\ of\ NH_3 = 30.3\ g$$