Find Remaining NO Concentration: 2NO + 2H2 → N2 + 2H2O, 0.006M NO, 0.002M H2

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Discussion Overview

The discussion revolves around calculating the remaining concentration of NO in the reaction 2NO + 2H2 → N2 + 2H2O, given initial concentrations of 0.006 M for NO and 0.002 M for H2, after half of the original amount of H2 has been consumed. Participants explore stoichiometric relationships and request clarification on how to approach similar problems with different reaction ratios.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant asks how to find the concentration of NO remaining after half of H2 has been consumed, expressing a desire for clarity on both the current reaction and a hypothetical non-1:1 reaction.
  • Another participant suggests assuming a 1L volume to simplify the stoichiometric calculations and prompts the original poster to determine how many moles of H2 reacted and how many moles of NO would react with that amount of H2.
  • A third post reiterates the original question and provides a calculation based on the stoichiometric ratio, concluding that 0.005 M of NO remains after the consumption of H2.
  • A fourth post reminds participants of forum rules regarding homework assistance and expresses concern over providing final answers to questions posed by others.

Areas of Agreement / Disagreement

There is no consensus on the appropriateness of providing direct answers to homework questions, as one participant challenges the practice while others engage in calculations. The discussion remains unresolved regarding the best approach to answering such queries.

Contextual Notes

Participants do not clarify the assumptions made regarding the volume of the reaction mixture or the implications of different reaction stoichiometries on the calculations.

Who May Find This Useful

Students or individuals seeking assistance with stoichiometry in chemical reactions, particularly in understanding concentration changes during reactions.

reyrey389
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2NO + 2H2 --> N2 + 2H2O

The concentration of NO is 0.006 M and the concentration for H2 is 0.002M. How can I find the concentration of NO remaining when one-half of the original amount of H2 has been consumed?

if you can also show me how it would be done if the reaction was not 1:1, that would help a lot. thanks
 
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It is a basic stoichiometry. To simplify things you can assume 1L volume. How many moles of H2 reacted if half was consumed? How many moles of NO reacted with this amount of hydrogen?
 
reyrey389 said:
2NO + 2H2 --> N2 + 2H2O

The concentration of NO is 0.006 M and the concentration for H2 is 0.002M. How can I find the concentration of NO remaining when one-half of the original amount of H2 has been consumed?

if you can also show me how it would be done if the reaction was not 1:1, that would help a lot. thanks
Here the molar ratio for NO and H2 is 1:1.

We know that 1/2 of original H2 remains
0.002/2 =0.001
for every 1 NO 1 H2 gets used up
for every 0.001 NO 0.001 H2 gets used up

0.006-0.001 =0.005 NO remaining
 
Have you read the forum rules? Please don't do homework for others. A week passed, but it is not a reason to give final answers.
 

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