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.

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