Below are the reduction half reactions for chemolithoautotrophic denitrification, where hydrogen is a source of electrons and energy and nitrate is the terminal electron acceptor. NO3 + 10e-> N₂ (E° = +0.74 V) • H+ + 2e -> H₂ (E° = -0.42 V) If you balance and combine the reactions so that 110 moles of H₂ gas are oxidized to H*, how many moles of NO3 will be reduced to dinitrogen gas?

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Below are the reduction half reactions for chemolithoautotrophic denitrification, where hydrogen is a source of electrons and energy and nitrate is the terminal electron
acceptor.
• NO3 + 10e-> N₂ (E°= +0.74 V)
• H+ + 2e -> H₂ (E° = -0.42 V)
If you balance and combine the reactions so that 110 moles of H₂ gas are oxidized to H*, how many moles of NO3 will be reduced to dinitrogen gas?
•
Report your answer in standard decimal notation rounded to two decimal places.
•
Include trailing zeros, even if the answer is a whole number
e.g. 18.00, not 18
• Report only the numeric part of your answer
o e.g. 18.00, not 18.00 moles NO3
Transcribed Image Text:Below are the reduction half reactions for chemolithoautotrophic denitrification, where hydrogen is a source of electrons and energy and nitrate is the terminal electron acceptor. • NO3 + 10e-> N₂ (E°= +0.74 V) • H+ + 2e -> H₂ (E° = -0.42 V) If you balance and combine the reactions so that 110 moles of H₂ gas are oxidized to H*, how many moles of NO3 will be reduced to dinitrogen gas? • Report your answer in standard decimal notation rounded to two decimal places. • Include trailing zeros, even if the answer is a whole number e.g. 18.00, not 18 • Report only the numeric part of your answer o e.g. 18.00, not 18.00 moles NO3
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