If the actual yield of the reaction was 75% instead of 100%, how many molecules of no would be present after the reaction was over? express your answer as an integer.

Question

Marisol Austin

Chemistry

25 April, 01:58

If the actual yield of the reaction was 75% instead of 100%, how many molecules of no would be present after the reaction was over? express your answer as an integer.

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Dylan Middleton · Answer 1

To be able to answer this equations, we must set given information. Suppose the reaction to yield NO is:

N₂ + O₂ → 2 NO

Next, suppose you have 1 g of each of the reactants. Determine first which is the limiting reactant.

1 g N₂ (1 mol N₂ / 28 g) (2 mol NO/1 mol N₂) = 0.07154 mol NO present

Number of molecules = 0.07154 mol NO (6.022*10²³ molecules/mol)

Number of molecules = 4.3*10²² molecules NO present