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For the reaction NO_2(g) + NO (g) ⇌ N_2O(g) + O_2(g), K_c = 0.857. Equimolar amounts...

For the reaction NO_2(g) + NO (g) ⇌ N_2O(g) + O_2(g), K_c = 0.857. Equimolar amounts of NO and NO_2 gases are to be introduced simultaneously into a 2.50 L flask until the concentration of N_2O exactly equals 0.0300 M. How many grams of NO gas must be initially introduced? Do not enter units as part of your answer. Report your answer to the correct number of significant figures.

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Answer #1

t=te Reaction: NO2(g) + NO(g) = N2069) + O2(9) t=0 a a o o a-x a-x x x as initial number of moles X = number of moles consume

So, initial moles of no must be 0.156 mol: Molar mass of No = 30 g molt - Grams of No needed = (0:156 x 30) - 4680 g 46

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