2 moles of H20 heated at 10A for 1hr. What is the volume?

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Discussion Overview

The discussion centers on the electrolysis of 2 moles of water at 10A for 1 hour, specifically focusing on calculating the total volume of the system after the process. The scope includes theoretical and mathematical reasoning related to electrolysis and gas volume calculations.

Discussion Character

  • Homework-related, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant presents an initial calculation for the final volume after electrolysis, suggesting a total of 39.358mL based on the mass of gases produced.
  • Another participant challenges the volume calculation, stating that 3.358g of gas does not equate to 3.358mL and points out the discrepancy in mass when gases are removed from water.
  • A third participant acknowledges their inexperience in chemistry and revises the final volume to 32.65mL, interpreting it as the remaining volume of water.
  • Several participants emphasize the necessity of knowing pressure and temperature to accurately determine the total volume of the system, suggesting that the question may be poorly worded or assumes standard temperature and pressure (STP).
  • One participant reiterates the importance of clarifying the total volume occupied versus the volume of water remaining.

Areas of Agreement / Disagreement

Participants express disagreement regarding the interpretation of total volume and the calculations involved. There is no consensus on the correct approach or final volume due to differing assumptions and interpretations of the question.

Contextual Notes

Limitations include the lack of information on pressure and temperature, which are critical for accurate volume calculations of gases produced during electrolysis. The discussion also highlights potential ambiguities in the wording of the original question.

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Homework Statement


2 moles of water undergoes electrolysis at 10A. What is the total volume of the system after 1hr?

Homework Equations


4 H+(aq) + 4e−→ 2H2(g)

2 H2O(l) → O2(g) + 4 H+(aq) + 4e−

Faraday's Law of Electrolysis
Q = n(e-) x F

F = 96500C

The Attempt at a Solution



Initial volume = 36mL

Gas produced
(10*3600s*4g)/(F*4)=0.373g H2

(10*3600s*32g)/(F*4)=2.984g O2

0.373g+2.984g=3.358g

Final volume
36mL+3.35mL =39.358mL
 
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3.358 rounds up to 3.36
3.358g of gas does not make 3.358mL of gas.
If you start with 36g of H2O and you remove some of the H and some of the O you do not end up with 36g.
 
I apologize. I'm very new to chemistry. I thought I was solving for total volume including the expanding gas.

Final volume
36mL-3.35mL =32.65
 
You can't calculate total volume of "the system" not knowing the pressure and the temperature. Perhaps it is just a matter of lousy wording of the question, or you are expected to assume STP.

What you have found is the volume of water left. This is not total volume occupied.
 
Borek said:
You can't calculate total volume of "the system" not knowing the pressure and the temperature. Perhaps it is just a matter of lousy wording of the question, or you are expected to assume STP.

What you have found is the volume of water left. This is not total volume occupied.

Yes, it was very poor wording. Still new to writing these equations. Thank you for your help.
 

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