Chemistry ? How many grams of NH3 must be reacted with excess oxygen to yield 125 g of NO

I just need to be guided into the right direction to solve this problem; not looking for the answer, just looking for a little help. (answer would be helpful though)
Under certain conditions the reaction below will proceed at 28.2% yield of NO. How many grams of NH3 must be reacted with excess oxygen to yield 125g of NO?

4 NH3 + 5 O2 → 4 NO + 6 H2O

(125 g NO) / (30.0061 g NO/mol) x (4/4) x (17.0306 g NH3/mol) / (0.282) = 252 g NH3