If 7.50 mol of NH₃ and 8.20 mol of O₂ react in the following reaction, how many moles of H₂O will be formed? 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g)

Answer 1

9.84 moles of water. You need to determine the limiting reagent. Using the molar coefficients in the balanced equation, 6 moles of water are formed for every 4 moles of NH₃, and 6 moles of water are formed for every 5 moles of O₂. So (7.5 moles NH₃)(6 moles H₂O)/4 moles NH₃)=11.25 moles water and for (8.20 moles O₂)/(6 moles H₂O)(5 moles O₂)=9.84 moles H₂O.. So O₂ is the limiting reagent and we can only make as much product as allowed by the limiting reagent.