Matthew C. answered 05/21/19
Bachelor's of Science in Chemistry
a) 35.0 g N2 x (1 mol N2/28 g N2) x (2 mol NO/1 mol N2) x (30 g NO/1 mol NO) =
75 g NO
b) 35.0 g O2 x (1 mol O2/32 g O2) x (2 mol NO/1 mol O2) x (30 g NO/1 mol NO) =
65.625 g NO
c) 65.625 g NO
d) Since we know how much NO is produced for 35.0 g of N2 and 35.0 g of O2, the true amount produced will be the lower amount, because of that reactant being the limiting reagent (which in this case, is O2). Therefore, 65.625 g NO will be produced. Another way to think about it is that there isn't enough O2 to react with the 35.0 g of N2 that would've given you 75 g NO.