If 97.3 l of no2 forms measured at 35 c and 632 mm hg. what is the percent yield?

a. Use PV = nRT and solve for n = number of mols O2. 
mols NO = grams/molar mass = ? 

 Using the coefficients in the balanced equation, convert mols O2 to mols NO2. Do the same for mols NO to mols NO2. It is likely that the two values will not be the same which means one is wrong; the correct value in LR (limiting reagent) problems is ALWAYS the smaller value and the reagent producing that value is the LR. 

b. 
Using the smaller value for mols NO2 from part a, substitute for n in PV = nRT, use the conditions listed in part b, and solve for V in liters. This will give you the theoretical yield (YY)in liters. The actual yield at these same conditions (AY) is 84.8 L. 

and % will be 60%.

The correct answer is "atomic weight".

0.50 hope this