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?
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
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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
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
To determine the number of kilojoules released when 92.3 g of NO2 reacts, we need to use the given DH value (-137 kJ) and the stoichiometry of the reaction.
1. Start by calculating the moles of NO2 using its molar mass:
Moles of NO2 = mass (g) / molar mass (g/mol)
The molar mass of NO2 is:
Molar mass of N = 14.01 g/mol
Molar mass of O = 16.00 g/mol (there are two oxygen atoms)
Total molar mass of NO2 = 14.01 g/mol + 2 * 16.00 g/mol = 46.01 g/mol
Moles of NO2 = 92.3 g / 46.01 g/mol
2. Now, using the stoichiometry of the reaction, we can determine the moles of HNO3 and NO formed when the given amount of NO2 reacts.
From the balanced equation, we can see that the ratio of NO2 to HNO3 is 3:2 and the ratio of NO2 to NO is 3:1.
Moles of HNO3 = (Moles of NO2) * (2/3)
Moles of NO = Moles of NO2
3. Finally, calculate the energy released using the equation:
Energy released (in kJ) = moles of NO2 * DH
Substitute the values calculated in steps 1 and 2 into the equation to get the final answer.