er...... From your reaction equation, 3 moles of hydrogen gas produces 2 moles of AMMONIA if the reaction was 100% efficient. From the way you phrased your question, it's difficult to tell if you mean 450g of hydrogen (H), or 450 GRAMS of hydrogen gas (H2). I'll assume you mean hydrogen gas. FIRST you have to calculate the number of moles hydrogen gas you started with: Moles H2 = grams H2 / Molecular weight H2 Moles H2= 450 g H2 / 2.016 g per mole H2 Moles H2 =223.21 moles If the reaction was 100% efficient: Moles NH3 = 2/3 * moles H2 Moles NH3 = 2/3 * 223.21 Moles NH3 = 148.80 moles Now to calculate the grams NH3 expected from 100% efficiency: grams NH3 = moles NH3 at 100% efficiency * mol wt NH3 grams NH3 = 148.80 moles * 17.024 grams/mole grams NH3 = 2533.17 grams So to calculate the percent yield for the reaction: % yield = (actual yield/ theoretical yield) *100 % yield = (1575 grams/ 2533.17 grams) *100 % yield = 62.17% Now if you meant H instead of H2 for the original grams hydrogen in your question, you'll have to recalculate as appropriate....Hope this will HELP u..