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?

3 answers

kJ released = 137 kJ x (92.3/(3*molar mass NO2) = ?
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
-
6.018 moles NO2 x 137 kJ / 3 moles NO2 = 2.75 x 10^2 kJ
answer: 2.75 x 10^2 kJ
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
-
2.000609 moles NO2 x 137 kJ / 3 moles NO2 = 91.6 kJ
Similar Questions
  1. I have a few question for my review.1) How would you prepare 30.0 ml of 0.600M HNO3 From a stock solution of 4.00M HNO3? What
    1. answers icon 4 answers
  2. Given the reaction3 NO2(g) + H2O(l) �¨ 2 HNO3(l) + NO(g) ƒ¢rH = -72.0 kJ, calculate the molar enthalpy of reaction, rH for:
    1. answers icon 0 answers
    1. answers icon 0 answers
    1. answers icon 0 answers
more similar questions