Based on this equation 4NH3+5O2=4NO+6H2O How many grams of NO can be produced by reacting 0.650 moles of O2 with 9.88 grams of NH3?

Molar mass of NH₃ = (14.0 + 1.0×3) g/mol = 17.0 g/mol
Molar mass of NO = (14.0 + 16.0) g/mol = 30.0 g/mol

4NH₃ + 5O₂ → 4NO + 6H₂O
OR: Mole ratio NH₃ : O₂ = 4 : 5

Initial no. of moles of NH₃ = (9.88 g) / (17.0 g/mol) = 0.581 mol
Initial no. of moles of O₂ = 0.650 mol

When O₂ completely reacts, NH₃ needed = (0.650 mol) × (4/5) = 0.520 mol < 0.581 mol
Hence, NH₃ is in excess, and O₂ completely reacts.

According to the above equation, mole ratio O₂ : NO = 5 : 4
No. of moles of O₂ reacted = 0.650 mol
No. of moles of NO produced = (0.650 mol) × (4/5) = 0.520 mol
Mass of NO produced = (0.520 mol) × (30.0 g/mol) = 15.6 g