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

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SUMMARY

The discussion focuses on calculating the remaining concentration of nitric oxide (NO) in the reaction 2NO + 2H2 → N2 + 2H2O, given initial concentrations of 0.006 M for NO and 0.002 M for H2. When half of the original amount of H2 (0.001 M) is consumed, the stoichiometric relationship indicates that an equivalent amount of NO (0.001 M) is also consumed. Thus, the remaining concentration of NO is calculated to be 0.005 M. The discussion also touches on the implications of non-1:1 stoichiometric ratios in similar reactions.

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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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