2NO (g) + O2(g)? 2NO2 (g) If 9.0 L of NO is combined with excess O2 at STP, what

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2NO(g) + O2(g)------->2NO2(g) One mole of ANY gas at STP occupies a volume of 22.4 L. 9 L NO x (1 mole NO / 22.4 L NO) = 0.40 moles NO According to the balanced equation, 2 moles of NO will produce 2 moles of NO2. 0.40 moles NO x (2 moles NO2 / 2 moles NO) = 0.40 moles NO2 0.40 moles NO2 x (22.4 L NO2 / 1 mole NO2) = 9 L NO2 That's the "long" way. The "short" way is to apply Avogadro's law which says that at constant P and T, moles and volume are proportional. Since the balanced equation shows that moles of NO and moles of NO2 are in a 1:1 ratio, then volume NO = volume NO2 = 9L.

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