Chemistry Homework- Stoichiometry: Volume and Particle Calculations

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SUMMARY

The discussion focuses on stoichiometry related to the decomposition of water (H2O) to produce oxygen (O2). The balanced equation is 2H2O → 2H2 + O2. Using dimensional analysis, the volume of oxygen produced from 5.67 x 10^24 molecules of water is calculated to be 105.488L when applying the mol-mol ratio of 1 mol O2 to 2 mol H2O. Without this ratio, the calculation yields 210.977L, highlighting the importance of stoichiometric relationships in chemical calculations.

PREREQUISITES
  • Understanding of stoichiometry and balanced chemical equations
  • Familiarity with Avogadro's number (6.02 x 10^23 molecules/mol)
  • Knowledge of gas laws, specifically the volume of gas at STP (22.4L)
  • Proficiency in dimensional analysis techniques
NEXT STEPS
  • Study the concept of molar ratios in chemical reactions
  • Learn about the Ideal Gas Law and its applications
  • Explore advanced stoichiometric calculations involving multiple reactants and products
  • Investigate the kinetic molecular theory and its implications for gas behavior
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Chemistry students, educators, and anyone seeking to enhance their understanding of stoichiometry and gas calculations in chemical reactions.

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Chemistry Homework-- Stoichiometry: Volume and Particle Calculations

This is only to check my work, and I may be updating this post with more problems. There are ten questions total, and this is NOT graded. Therefore, the help I receive goes toward my understanding, not my score.

Homework Statement


What VOLUME of oxygen is produced from the decomposition of 5.67 x 10^24 molecules of water?
Balanced equation:
2H2O >decomposes to> 2H2 + O2


Homework Equations


6.02 x 10^23 molecules : 22.4L @STP for any gas
Mol-Mol ratio = 1 mol O2/ 2 mol H2O

The Attempt at a Solution


Trying to go along with dimensional analysis format.
(5.67 x 10^24)/(6.02 x 10^23) x 1 mol O2/ 2 mol H2O x 22.4L = 105.488L

Is the mol-mol ratio necessary? Without it, the answer comes out to 210.977L (rounded).
 
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SpoChi said:
This is only to check my work, and I may be updating this post with more problems. There are ten questions total, and this is NOT graded. Therefore, the help I receive goes toward my understanding, not my score.

Homework Statement


What VOLUME of oxygen is produced from the decomposition of 5.67 x 10^24 molecules of water?
Balanced equation:
2H2O >decomposes to> 2H2 + O2

Homework Equations


6.02 x 10^23 molecules : 22.4L @STP for any gas
Mol-Mol ratio = 1 mol O2/ 2 mol H2O

The Attempt at a Solution


Trying to go along with dimensional analysis format.
(5.67 x 10^24)/() x 1 mol O2/ 2 mol H2O x 22.4L = 105.488L

Is the mol-mol ratio necessary? Without it, the answer comes out to 210.977L (rounded).

Your answer is correct, or more cautiously, I get the same.
Why are you insecure?
A mole of gas occupies i.e. 6.02 x 1023 molecules of gas, occupies 22.4 l at standard atmospheric pressure and temperature.
Independently of how many atoms are in each molecule.
I.e. 6.02 x 1023 molecules of gas occupies 22.4 l under standard conditions - however many atoms there are in each molecule.

6.02 x 1023 molecules of Argon (A), 6.02 x 1023 molecules of oxygen (O2) or 6.02 x 1023molecules of ethane (C2H6) all occupy 22.4 l even though this is 6.02 x 1023 atoms of A, 2X6.02 x 1023 atoms of O2, or 8X6.02 x1023 atoms of ethane .

The rough basic reason for that is that any gas at a given pressure and temperature has the same average kinetic energy in the translational movement of the molecules. If it didn't then say you had a balloon of argon surrounded by oxygen at the same pressure then the argon atoms bounding of the inside of the balloon would transfer different momentum to it outwards than the oxygen atoms bouncing would transfer inwards. The balloon would expand or contract, in fact the pressure could not have been the same outside and inside to start with. I am not saying that is quite obvious, but that is what is shown in kinetic theory somewhere in your textbooks or maybe someone could explain better here.
 
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