Final answer:
The mass of NO formed from the given quantities of NH₃ and O₂ is calculated to be 44.05 grams, using stoichiometry and the conservation of mass principle. The closest answer choice with three significant figures would be 45.0 g, which matches the known products from similar reactions. There seems to be an error, as none of the provided answer options exactly match the calculated mass.
Step-by-step explanation:
To determine the mass of NO that will be formed from the reaction between 25.0 g of NH₃ and 42.7 g of O₂, we will first use stoichiometry to find the limiting reactant and then calculate the mass of NO formed.
First, we should convert the mass of NH₃ and O₂ to moles using their molar masses (NH₃ has a molar mass of 17.03 g/mol, and O₂ has a molar mass of 32.00 g/mol).
- For NH₃: 25.0 g NH₃ × (1 mol NH₃ / 17.03 g NH₃) = 1.468 moles NH₃
- For O₂: 42.7 g O₂ × (1 mol O₂ / 32.00 g O₂) = 1.334 moles O₂
The balanced chemical equation is 4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g). From this, we can see that 4 moles of NH₃ react with 5 moles of O₂. By comparing the mole ratio of NH₃ to O₂ that we have, the limiting reactant is NH₃ since we require more moles of O₂ than what we have for a complete reaction.
Using the molar ratio from the balanced equation (4 moles of NH₃ : 4 moles of NO), we can calculate the moles of NO that would be formed:
- 1.468 moles NH₃ × (4 moles NO / 4 moles NH₃) = 1.468 moles NO
Finally, we convert the moles of NO to grams:
- 1.468 moles NO × (30.01 g NO / 1 mol NO) = 44.05 g NO
The mass of NO formed is 44.05 grams, which is not an exact match to any of the answer choices provided, which suggests there may have been a typo or calculation error in the question or answer choices because each answer should have three significant figures, similar to the total mass of the products in the law of conservation of mass example provided. However, the principle of mass conservation requires that the answer be consistent in significant figures with the masses of reactants provided.