The combustion of 43.9 g of ammonia with 258 g of oxygen produces g of no2.

The balanced equation would be 4NH3 + 7O2 --> 4NO2 + 6H2O. Mass of Ammonia = 43.9 g; Molecular weight of Ammonia NH3 = 17.03056 g/mol. Mass of Oxygen = 258 g; Molecular weight of Oxygen O2 = 31.998 g/mol. Calculating the moles of Ammonia NH3 = Mass / Molecular weight of ammonia = 43.9 g / 17.03 g/mol = 2.578 moles of Ammonia. Calculating the moles of Oxygen O2 = Mass / Molecular weight of Oxygen =258 g / 31.998 g/mol = 8.063 moles of Oxygen. 2.578 moles of Ammonia will react with (7/4) x 2.578 moles of Oxygen (from the equation) that is 4.5115 moles of O2 to produce the NO2. Since we have more moles of O2 8.063 moles, Ammonia will be the limiting agent. Calculating the NO2 produces using the coordinate from the equation Molecular weight of NO2 = 46.00558 g/mol Mass of NO2 = (2.578 mol) x (4 mol of NO2 / 4 mol of NH3) x (46.006 g /mol) Produced NO2 = 118.6 g NO2

Ithinks its false. tell me if im wrong