yeah so you have to start of with converting your first two values into moles (forget the third one)
97.5 g NO * 1 mol/30.01 g NO = 3.25 moles NO
88.0 g O2 * 1 mol/16.00 g O2 = 5.5 moles O2
now we can find the limiting reactant by checking for the amount of product each reactant should give us by using molar ratios
3.25 mol NO * 2 mol NO2/2 mol NO = 3.25 mol NO2
5.5 mol O2 * 2 mol NO2/ 1 mol O2 = 11
so NO is the limiting reactant since it produces less product/gets used up quicker
3.25 mol NO * 2 mol NO2/2molNO = 3.25 mol NO2
so this is our theoretical yield and the question provides us with the actual yield (2.68 moles). since the actual yield is given in moles, we don't have to convert to grams. our percent yield formula goes like: actual yield/theoretical yield * 100
2.68 mol/3.25 mol * 100 = 82.46%