Question

Consider the reaction, 3 NO2(g) + H2O(l) ® 2 HNO3(aq) + NO(g),
where DH = – 137 kJ. How many kilojoules are released when 92.3 g of NO2 reacts?

Answer 1

kJ released = 137 kJ x (92.3/(3*molar mass NO2) = ?

Answer 2

Oh Lord, the first reply like...that was not the question, seriously?
NO2 mass: 46.01
DH -137 kJ
92.3g NO2 x 3 moles NO2 / 46.01 g NO2 = 6.018 moles
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6.018 moles NO2 x 137 kJ / 3 moles NO2 = 2.75 x 10^2 kJ
answer: 2.75 x 10^2 kJ

Answer 3

made a mistake lol divided by 3 moles when it should have been 1 in the first equation
NO2 mass: 46.01
DH -137 kJ
92.3g NO2 x 1 moles NO2 / 46.01 g NO2 = 2.000609 moles
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2.000609 moles NO2 x 137 kJ / 3 moles NO2 = 91.6 kJ