8.85 Part C

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Shannon Wasley 2J
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8.85 Part C

Postby Shannon Wasley 2J » Sat Feb 10, 2018 9:00 pm

The oxidation of nitrogen in the hot exhaust of jet engines and automobile engines occurs by the reaction
N2 (g) + O2 (g) 2 NO (g) ΔHo = +180.6 kJ

Part C: When the oxidation of N2 to NO was completed in a bomb calorimeter, the heat absorbed was measured as 492 J. What mass of nitrogen gas was oxidized?

For this part of the question, why do we simply divide .492 by 180.6 to get the moles of nitrogen gas?

Nisarg Shah 1C
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Joined: Sat Jul 22, 2017 3:00 am

Re: 8.85 Part C

Postby Nisarg Shah 1C » Sat Feb 10, 2018 9:05 pm

For every mole of N2 oxidized, the reaction absorbs 180.6kJ of heat. Therefore, if an unknown number of moles of N2 oxidized releases 492J, we can use divide this value by the molar enthalpy of this reaction (180.6kJ) to get the number of moles of N2 oxidized.


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