Question

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.

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

Answer 2

The reaction will be

2NO  + O2 ---< 2NO2

so as per balanced equation two moles of NO will react with one mole of O2 to give two moles of NO2

In this particular question, 5 molecules of O2 and 8 molecules of NO are present

so the limiting reagent is NO

now the 8 molecules of NO should react with 4 molecules of O2 for 100% yield

As the mentioned yield is 75% only, it means only 75% of NO molecules will react

75% of 8 = 6

So six molecules of NO will react with 3 molecules of O2 to give 75% yield

so after reaction we will be left with  2 molecules of NO