Final answer:
To find the mass of O₂ consumed in the combustion of 28.8 g of NH₃, use the stoichiometric ratio from the balanced chemical equation 4NH₃ + 3O₂ → 2N₂ + 6H₂O. This calculation results in a required mass of 40.64 g of O₂.
Step-by-step explanation:
To calculate the mass of O₂ consumed by the combustion of 28.8 g of NH₃, we first need a balanced chemical equation for the combustion of ammonia, which is typically:
4NH₃ + 3O₂ → 2N₂ + 6H₂O
Using the molar masses, we know that the molar mass of NH₃ is approximately 17 g/mol, and the molar mass of O₂ is 32 g/mol. From the balanced equation, we see that 4 moles of NH₃ react with 3 moles of O₂. This ratio (4:3 NH₃:O₂) allows us to set up a proportion to find the mass of O₂ needed for 28.8 g of NH₃. Since 1 mole of NH₃ is 17 g, 28.8 g of NH₃ is (28.8 g)/(17 g/mol) = 1.69 moles of NH₃. Using the stoichiometric ratio we find that the moles of O₂ needed are (3/4) × 1.69 moles = 1.27 moles O₂. The mass of O₂ required is then 1.27 moles × 32 g/mol = 40.64 g of O₂.
First, calculate the moles of O₂: 28.8 g NH₃ × (1 mol NH₃ / 17 g NH₃) × (5 mol O₂ / 4 mol NH₃) = 8.82 mol O₂
Finally, convert the moles of O₂ to grams: 8.82 mol O₂ × (32 g O₂ / 1 mol O₂) = 282 g O₂