Final answer:
The combustion of 43.9 grams of ammonia with 258 grams of oxygen produces 118.92 grams of NO₂. Thus, thereis no correct option.
Step-by-step explanation:
To determine the grams of NO₂ produced in the combustion of ammonia and oxygen, we first need to write a balanced equation for the reaction:
4NH₃ + 5O₂ -> 4NO₂ + 6H₂O
The molar mass of ammonia (NH₃) is 17.03 g/mol, and the molar mass of oxygen (O₂) is 32.00 g/mol. We can use these molar masses to convert the given masses of ammonia and oxygen into moles:
43.9 g NH₃ * (1 mol NH₃ / 17.03 g NH₃) = 2.58 mol NH₃
258 g O₂ * (1 mol O₂ / 32.00 g O₂) = 8.06 mol O₂
The balanced equation tells us that the ratio of NH₃ to NO₂ is 4:4. Therefore, the moles of NO₂ produced will be the same as the moles of NH₃ consumed:
2.58 mol NH₃ * (4 mol NO₂ / 4 mol NH₃) = 2.58 mol NO₂
Now, we can use the molar mass of NO₂ (46.01 g/mol) to convert the moles of NO₂ into grams:
2.58 mol NO₂ * (46.01 g NO₂ / 1 mol NO₂) = 118.92 g NO₂
Therefore, the combustion of 43.9 grams of ammonia with 258 grams of oxygen produces 118.92 grams of NO₂.
Thus, there is no correct option.