Suppose A 500 ML 500 \, \text{mL} 500 ML Flask Is Filled With 1.3 Mol 1.3 \, \text{mol} 1.3 Mol Of C H 4 CH_4 C H 4 ​ , 0.70 Mol 0.70 \, \text{mol} 0.70 Mol Of H 2 S H_2S H 2 ​ S , And 1.1 Mol 1.1 \, \text{mol} 1.1 Mol Of H 2 H_2 H 2 ​ . The Following Reaction Occurs:$[

Suppose a $500 \, \text{mL}$ flask is filled with $1.3 \, \text{mol}$ of $CH_4$, $0.70 \, \text{mol}$ of $H_2S$, and $1.1 \, \text{mol}$ of $H_2$. The following reaction occurs:

Introduction

Chemical reactions are a fundamental aspect of chemistry, and understanding the stoichiometry of these reactions is crucial in determining the amount of reactants and products involved. In this article, we will explore a specific chemical reaction involving methane ($CH_4$), hydrogen sulfide ($H_2S$), and hydrogen ($H_2$). We will analyze the given reaction and determine the limiting reactant, the amount of products formed, and the remaining reactants.

The Reaction

The reaction that occurs in the flask is:

$$\begin{aligned} CH_4 + H_2S &\rightarrow \text{C} + 2H_2 + 2H_2S \\ \end{aligned}$$

This reaction involves the combustion of methane in the presence of hydrogen sulfide, resulting in the formation of carbon, hydrogen gas, and hydrogen sulfide gas.

Stoichiometry of the Reaction

To determine the limiting reactant, we need to calculate the number of moles of each reactant required to produce the products. The balanced equation shows that 1 mole of $CH_4$ reacts with 1 mole of $H_2S$ to produce 1 mole of $C$, 2 moles of $H_2$, and 2 moles of $H_2S$.

Calculating the Number of Moles of Each Reactant

We are given the following initial amounts of reactants:

• $CH_4$: $1.3 \, \text{mol}$
• $H_2S$: $0.70 \, \text{mol}$
• $H_2$: $1.1 \, \text{mol}$

Since $H_2$ is not involved in the reaction, we can ignore it for now.

Determining the Limiting Reactant

To determine the limiting reactant, we need to calculate the number of moles of $CH_4$ and $H_2S$ required to produce the products.

Let's assume that $x$ moles of $CH_4$ react with $x$ moles of $H_2S$ to produce $x$ moles of $C$, $2x$ moles of $H_2$, and $2x$ moles of $H_2S$.

Since we are given the initial amounts of reactants, we can set up the following equations:

$$\begin{aligned} x &= 1.3 \, \text{mol} \\ x &= 0.70 \, \text{mol} \\ \end{aligned}$$

Solving these equations, we get:

$$\begin{aligned} x &= 0.70 \, \text{mol} \\ \end{aligned}$$

This means that $0.70 \, \text{mol}$ of $CH_4$ and $0.70 \, \text{mol}$ of $H_2S$ are required to produce the products.

Calculating the Amount of Products Formed

Since $0.70 \, \text{mol}$ of $CH_4$ and $0.70 \, \text{mol}$ of $H_2S$ react to produce $0.70 \, \text{mol}$ of $C$, $1.4 \, \text{mol}$ of $H_2$, and $1.4 \, \text{mol}$ of $H_2S$, we can calculate the amount of products formed as follows:

• $C$: $0.70 \, \text{mol}$
• $H_2$: $1.4 \, \text{mol}$
• $H_2S$: $1.4 \, \text{mol}$

Calculating the Remaining Reactants

Since $0.70 \, \text{mol}$ of $CH_4$ and $0.70 \, \text{mol}$ of $H_2S$ react, the remaining amounts of reactants are:

• $CH_4$: $1.3 \, \text{mol} - 0.70 \, \text{mol} = 0.60 \, \text{mol}$
• $H_2S$: $0.70 \, \text{mol} - 0.70 \, \text{mol} = 0 \, \text{mol}$

Conclusion

In this article, we analyzed a specific chemical reaction involving methane, hydrogen sulfide, and hydrogen. We determined the limiting reactant, the amount of products formed, and the remaining reactants. The limiting reactant was found to be $H_2S$, and the amount of products formed was $0.70 \, \text{mol}$ of $C$, $1.4 \, \text{mol}$ of $H_2$, and $1.4 \, \text{mol}$ of $H_2S$. The remaining reactants were $0.60 \, \text{mol}$ of $CH_4$ and $0 \, \text{mol}$ of $H_2S$.

References

• [1] Petrucci, R. H., Harwood, W. S., Herring, F. G., & Madura, J. D. (2016). General chemistry: Principles and modern applications. Pearson Education.
• [2] Atkins, P. W., & De Paula, J. (2010). Physical chemistry. Oxford University Press.

Further Reading

• [1] Stoichiometry of chemical reactions
• [2] Limiting reactants
• [3] Chemical equilibrium
  Suppose a $500 \, \text{mL}$ flask is filled with $1.3 \, \text{mol}$ of $CH_4$, $0.70 \, \text{mol}$ of $H_2S$, and $1.1 \, \text{mol}$ of $H_2$. The following reaction occurs:

Q&A

Q: What is the limiting reactant in the given reaction?

A: The limiting reactant is $H_2S$, as it is the reactant that is consumed first in the reaction.

Q: How many moles of $C$, $H_2$, and $H_2S$ are formed in the reaction?

A: $0.70 \, \text{mol}$ of $C$, $1.4 \, \text{mol}$ of $H_2$, and $1.4 \, \text{mol}$ of $H_2S$ are formed in the reaction.

Q: What is the remaining amount of $CH_4$ and $H_2S$ after the reaction?

A: The remaining amount of $CH_4$ is $0.60 \, \text{mol}$, and the remaining amount of $H_2S$ is $0 \, \text{mol}$.

Q: Why is $H_2$ not involved in the reaction?

A: $H_2$ is not involved in the reaction because it is not a reactant in the given reaction.

Q: What is the purpose of the reaction?

A: The purpose of the reaction is to produce $C$, $H_2$, and $H_2S$ from $CH_4$ and $H_2S$.

Q: What are the conditions required for the reaction to occur?

A: The conditions required for the reaction to occur are not specified in the problem, but it is assumed that the reaction occurs under standard conditions.

Q: What are the products of the reaction?

A: The products of the reaction are $C$, $H_2$, and $H_2S$.

Q: What is the stoichiometry of the reaction?

A: The stoichiometry of the reaction is 1:1:2, meaning that 1 mole of $CH_4$ reacts with 1 mole of $H_2S$ to produce 1 mole of $C$, 2 moles of $H_2$, and 2 moles of $H_2S$.

Additional Questions

• What is the molar ratio of $CH_4$ to $H_2S$ in the reaction?
• What is the molar ratio of $H_2$ to $H_2S$ in the reaction?
• What is the molar ratio of $C$ to $H_2$ in the reaction?
• What is the molar ratio of $C$ to $H_2S$ in the reaction?

Answers

• The molar ratio of $CH_4$ to $H_2S$ is 1:1.
• The molar ratio of $H_2$ to $H_2S$ is 1:1.
• The molar ratio of $C$ to $H_2$ is 1:2.
• The molar ratio of $C$ to $H_2S$ is 1:2.

Conclusion

In this article, we have answered various questions related to the given reaction. We have determined the limiting reactant, the amount of products formed, and the remaining reactants. We have also discussed the stoichiometry of the reaction and the conditions required for the reaction to occur.