The Following Balanced Equation Shows The Formation Of Water:${ 2 H_2 + O_2 \rightarrow 2 H_2O }$How Many Moles Of Oxygen { (O_2)$}$ Are Required To Completely React With 27.4 Mol Of { H_2$}$?A. 6.8 Mol B. 13.7 Mol C.

Mar 10, 2025 by ADMIN 221 views

The Formation of Water: Balancing Chemical Equations and Stoichiometry

Understanding the Balanced Equation

The balanced chemical equation for the formation of water is given as:

${ 2 H_2 + O_2 \rightarrow 2 H_2O }$

This equation indicates that 2 moles of hydrogen gas ( $H_2$ ) react with 1 mole of oxygen gas ( $O_2$ ) to produce 2 moles of water ( $H_2O$ ). In this article, we will focus on the stoichiometry of this reaction and determine the number of moles of oxygen required to completely react with a given amount of hydrogen.

Stoichiometry and Mole Ratios

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. In this case, we are interested in the mole ratio between hydrogen and oxygen. According to the balanced equation, 2 moles of $H_2$ react with 1 mole of $O_2$ . This means that the mole ratio of $H_2$ to $O_2$ is 2:1.

Calculating the Number of Moles of Oxygen Required

We are given that 27.4 mol of $H_2$ are available for reaction. To determine the number of moles of $O_2$ required, we can use the mole ratio from the balanced equation. Since 2 moles of $H_2$ react with 1 mole of $O_2$ , we can set up a proportion to relate the number of moles of $H_2$ to the number of moles of $O_2$ :

${ \frac{2 \text{ mol } H_2}{1 \text{ mol } O_2} = \frac{27.4 \text{ mol } H_2}{x \text{ mol } O_2} }$

To solve for $x$ , we can cross-multiply and divide:

${ 2x = 27.4 }$

${ x = \frac{27.4}{2} }$

${ x = 13.7 \text{ mol } O_2 }$

Therefore, 13.7 mol of $O_2$ are required to completely react with 27.4 mol of $H_2$ .

Conclusion

In this article, we have used the balanced chemical equation for the formation of water to determine the number of moles of oxygen required to completely react with a given amount of hydrogen. By applying the principles of stoichiometry and mole ratios, we have shown that 13.7 mol of $O_2$ are required to react with 27.4 mol of $H_2$ . This calculation demonstrates the importance of understanding the quantitative relationships between reactants and products in chemical reactions.

Additional Examples and Applications

The principles of stoichiometry and mole ratios can be applied to a wide range of chemical reactions and problems. Some additional examples and applications include:

Calculating the number of moles of a product required to react with a given amount of a reactant
Determining the limiting reactant in a chemical reaction
Calculating the yield of a chemical reaction
Designing chemical reactors and processes

These examples and applications demonstrate the importance of understanding stoichiometry and mole ratios in chemistry and their relevance to real-world problems and applications.

Common Mistakes and Misconceptions

There are several common mistakes and misconceptions that can occur when working with stoichiometry and mole ratios. Some of these include:

Failing to balance the chemical equation
Misinterpreting the mole ratio between reactants and products
Failing to account for the stoichiometry of the reaction
Making errors in calculation or unit conversion

These mistakes and misconceptions can lead to incorrect conclusions and results. It is essential to carefully read and understand the problem, balance the chemical equation, and apply the principles of stoichiometry and mole ratios correctly.

Real-World Applications of Stoichiometry

Stoichiometry and mole ratios have numerous real-world applications in fields such as:

Chemical engineering: Stoichiometry is used to design and optimize chemical reactors and processes.
Materials science: Stoichiometry is used to determine the composition and properties of materials.
Environmental science: Stoichiometry is used to understand and mitigate the environmental impacts of chemical reactions.
Biotechnology: Stoichiometry is used to design and optimize biotechnological processes.

These applications demonstrate the importance of understanding stoichiometry and mole ratios in real-world problems and applications.

Conclusion

In conclusion, the balanced chemical equation for the formation of water is a fundamental concept in chemistry that can be used to determine the number of moles of oxygen required to completely react with a given amount of hydrogen. By applying the principles of stoichiometry and mole ratios, we have shown that 13.7 mol of $O_2$ are required to react with 27.4 mol of $H_2$ . This calculation demonstrates the importance of understanding the quantitative relationships between reactants and products in chemical reactions.
The Formation of Water: Balancing Chemical Equations and Stoichiometry - Q&A

Frequently Asked Questions

In this article, we will address some of the most common questions and concerns related to the formation of water, balancing chemical equations, and stoichiometry.

Q: What is the balanced chemical equation for the formation of water?

A: The balanced chemical equation for the formation of water is:

${ 2 H_2 + O_2 \rightarrow 2 H_2O }$

Q: What is the mole ratio between hydrogen and oxygen in this reaction?

A: The mole ratio between hydrogen and oxygen in this reaction is 2:1. This means that 2 moles of hydrogen gas ( $H_2$ ) react with 1 mole of oxygen gas ( $O_2$ ) to produce 2 moles of water ( $H_2O$ ).

Q: How many moles of oxygen are required to completely react with 27.4 mol of hydrogen?

A: To determine the number of moles of oxygen required, we can use the mole ratio from the balanced equation. Since 2 moles of $H_2$ react with 1 mole of $O_2$ , we can set up a proportion to relate the number of moles of $H_2$ to the number of moles of $O_2$ :

${ \frac{2 \text{ mol } H_2}{1 \text{ mol } O_2} = \frac{27.4 \text{ mol } H_2}{x \text{ mol } O_2} }$

To solve for $x$ , we can cross-multiply and divide:

${ 2x = 27.4 }$

${ x = \frac{27.4}{2} }$

${ x = 13.7 \text{ mol } O_2 }$

Therefore, 13.7 mol of $O_2$ are required to completely react with 27.4 mol of $H_2$ .

Q: What is the limiting reactant in this reaction?

A: In this reaction, hydrogen is the limiting reactant. This means that the amount of hydrogen available for reaction is less than the amount of oxygen required to completely react with it.

Q: How can I determine the limiting reactant in a chemical reaction?

A: To determine the limiting reactant in a chemical reaction, you can use the mole ratio from the balanced equation and the amounts of reactants available. If the amount of one reactant is less than the amount required to completely react with the other reactant, then that reactant is the limiting reactant.

Q: What is the yield of this reaction?

A: The yield of this reaction is 2 moles of water ( $H_2O$ ) for every 2 moles of hydrogen gas ( $H_2$ ) and 1 mole of oxygen gas ( $O_2$ ) that react.

Q: How can I calculate the yield of a chemical reaction?

A: To calculate the yield of a chemical reaction, you can use the mole ratio from the balanced equation and the amounts of reactants available. The yield is the amount of product formed per unit amount of reactant.

Q: What are some common mistakes and misconceptions related to stoichiometry and mole ratios?

A: Some common mistakes and misconceptions related to stoichiometry and mole ratios include:

Failing to balance the chemical equation
Misinterpreting the mole ratio between reactants and products
Failing to account for the stoichiometry of the reaction
Making errors in calculation or unit conversion

Q: What are some real-world applications of stoichiometry and mole ratios?

A: Stoichiometry and mole ratios have numerous real-world applications in fields such as:

Chemical engineering: Stoichiometry is used to design and optimize chemical reactors and processes.
Materials science: Stoichiometry is used to determine the composition and properties of materials.
Environmental science: Stoichiometry is used to understand and mitigate the environmental impacts of chemical reactions.
Biotechnology: Stoichiometry is used to design and optimize biotechnological processes.

These applications demonstrate the importance of understanding stoichiometry and mole ratios in real-world problems and applications.

Conclusion