in one experiment .886 mole of NO is mixed with .503 mole of O2.calculate which

Question

in one experiment .886 mole of NO is mixed with .503 mole of O2.calculate which of the two reactants is the limiting reagent.calculate also the number of moles on NO2 produced. Chang edition 9th chapter 3 problem # 83 need help thank you
Chang edition 9th chapter 3 problem # 83 need help thank you

Explanation / Answer

The equation for the reaction is : 2 NO + O2 -----> 2 NO2 . Two moles of NO can react with one mole of O2 So 0.503 moles of O2 require 2* 0.503 = 1.006 moles of NO Since we have only 0.886 moles of NO, NO is the limitingreagent. . Also 0.886 moles of NO require 0.886 /2 = 0.443 moles of O2 So we have excess of O2. . So 0.886 moles of NO will react with 0.443 moles of O2 to give0.886 moles of NO2.

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