What Mass Of Oxygen { \left( O_2 \right)$}$ Forms In A Reaction That Produces ${ 15.90 \, \text{g} \, C_6H_{12}O_6\$} ?- Molar Mass Of { O_2 = 32.00 , \text{g/mol}$} − M O L A R M A S S O F \[ - Molar Mass Of \[ − M O L A R Ma Sso F \[ C_6H_{12}O_6 = 180.18 ,
Understanding the Problem
In this problem, we are tasked with determining the mass of oxygen that forms in a reaction that produces 15.90 g of glucose (C6H12O6). To solve this problem, we need to understand the chemical equation for the production of glucose and the molar masses of the reactants and products involved.
Chemical Equation for Glucose Production
The chemical equation for the production of glucose is:
C6H12O6 → 6CO2 + 6H2O
However, since we are interested in the production of oxygen (O2), we need to consider the balanced chemical equation for the combustion of glucose:
C6H12O6 + 6O2 → 6CO2 + 6H2O
Molar Mass of Reactants and Products
The molar masses of the reactants and products involved in the reaction are:
- Molar mass of O2 = 32.00 g/mol
- Molar mass of C6H12O6 = 180.18 g/mol
Calculating the Mass of Oxygen
To calculate the mass of oxygen that forms in the reaction, we need to first calculate the number of moles of glucose produced. We can do this by dividing the mass of glucose produced (15.90 g) by the molar mass of glucose (180.18 g/mol):
moles of glucose = mass of glucose / molar mass of glucose = 15.90 g / 180.18 g/mol = 0.0882 mol
Since the balanced chemical equation shows that 6 moles of O2 are produced for every 1 mole of glucose, we can calculate the number of moles of O2 produced:
moles of O2 = 6 × moles of glucose = 6 × 0.0882 mol = 0.5292 mol
Now, we can calculate the mass of O2 produced by multiplying the number of moles of O2 by the molar mass of O2:
mass of O2 = moles of O2 × molar mass of O2 = 0.5292 mol × 32.00 g/mol = 16.97 g
Therefore, the mass of oxygen that forms in a reaction that produces 15.90 g of glucose is approximately 16.97 g.
Conclusion
In this problem, we used the balanced chemical equation for the combustion of glucose to determine the mass of oxygen that forms in a reaction that produces 15.90 g of glucose. We calculated the number of moles of glucose produced and then used the mole ratio from the balanced chemical equation to calculate the number of moles of O2 produced. Finally, we calculated the mass of O2 produced by multiplying the number of moles of O2 by the molar mass of O2. The result shows that the mass of oxygen that forms in the reaction is approximately 16.97 g.
Additional Information
- The molar mass of O2 is 32.00 g/mol.
- The molar mass of C6H12O6 is 180.18 g/mol.
- The balanced chemical equation for the combustion of glucose is: C6H12O6 + 6O2 → 6CO2 + 6H2O.
References
- [1] Chemistry: An Atoms First Approach. By Steven S. Zumdahl. 3rd edition. Houghton Mifflin Company. 2005.
- [2] General Chemistry: Principles and Modern Applications. By Linus Pauling. 9th edition. W.H. Freeman and Company. 1988.
Frequently Asked Questions (FAQs) About the Mass of Oxygen in a Reaction Producing Glucose =====================================================================================
Q: What is the chemical equation for the production of glucose?
A: The chemical equation for the production of glucose is:
C6H12O6 → 6CO2 + 6H2O
However, since we are interested in the production of oxygen (O2), we need to consider the balanced chemical equation for the combustion of glucose:
C6H12O6 + 6O2 → 6CO2 + 6H2O
Q: What is the molar mass of oxygen (O2)?
A: The molar mass of oxygen (O2) is 32.00 g/mol.
Q: What is the molar mass of glucose (C6H12O6)?
A: The molar mass of glucose (C6H12O6) is 180.18 g/mol.
Q: How do I calculate the mass of oxygen that forms in a reaction producing glucose?
A: To calculate the mass of oxygen that forms in a reaction producing glucose, you need to follow these steps:
- Calculate the number of moles of glucose produced by dividing the mass of glucose produced by the molar mass of glucose.
- Use the mole ratio from the balanced chemical equation to calculate the number of moles of O2 produced.
- Multiply the number of moles of O2 by the molar mass of O2 to calculate the mass of O2 produced.
Q: What is the mass of oxygen that forms in a reaction producing 15.90 g of glucose?
A: The mass of oxygen that forms in a reaction producing 15.90 g of glucose is approximately 16.97 g.
Q: What is the significance of the balanced chemical equation in this problem?
A: The balanced chemical equation is crucial in this problem because it shows the mole ratio between glucose and oxygen. This mole ratio is used to calculate the number of moles of O2 produced, which is then used to calculate the mass of O2 produced.
Q: Can I use this method to calculate the mass of oxygen that forms in a reaction producing any other substance?
A: Yes, you can use this method to calculate the mass of oxygen that forms in a reaction producing any other substance, provided you have the balanced chemical equation and the molar masses of the reactants and products involved.
Q: What are some common mistakes to avoid when calculating the mass of oxygen that forms in a reaction?
A: Some common mistakes to avoid when calculating the mass of oxygen that forms in a reaction include:
- Not using the balanced chemical equation to determine the mole ratio between the reactants and products.
- Not calculating the number of moles of the reactant and product correctly.
- Not multiplying the number of moles of the product by the molar mass of the product to calculate the mass of the product.
Q: How can I apply this knowledge to real-world problems?
A: You can apply this knowledge to real-world problems by using the principles of stoichiometry to calculate the mass of oxygen that forms in a reaction producing any substance. This can be useful in a variety of fields, including chemistry, biology, and environmental science.
Conclusion
In this article, we have answered some frequently asked questions about the mass of oxygen that forms in a reaction producing glucose. We have also provided a step-by-step guide on how to calculate the mass of oxygen that forms in a reaction producing any substance. By following these steps and using the principles of stoichiometry, you can apply this knowledge to real-world problems and make informed decisions in a variety of fields.