Final answer:
Through stoichiometric calculations, it's determined that ammonia (NH3) is actually the limiting reactant, not the excess reactant as initially indicated. Therefore, there will be no excess NH3 remaining. Also, there will be no excess O2 as all of it will react with NH3.
Step-by-step explanation:
To determine how much excess reactant remains after the reaction between ammonia (NH3) and oxygen (O2), we need to perform stoichiometric calculations based on the balanced chemical equation 4NH3 + 5O2 → 4NO + 6H2O. First, we will find out the limiting reactant by comparing the molar amounts of each reactant with their stoichiometric coefficients.
From the molar masses of NH3 and O2, we calculate that 2.00 g of NH3 is equivalent to 2.00 g / 17.03 g/mol = 0.117 mol of NH3, and 4.00 g of O2 is equivalent to 4.00 g / 32.00 g/mol = 0.125 mol of O2. According to the stoichiometry of the balanced equation, 4 mol of NH3 reacts with 5 mol of O2. This gives us a molar ratio of NH3 to O2 as 4:5. Using the actual amounts in moles, the comparative ratio is 0.117 mol NH3 to 0.125 mol O2, which is roughly 0.94:1.
Because the ratio of NH3 to O2 in the reaction should be 4:5, and we have less NH3 than required by the ratio with the given amounts (0.94:1 instead of 4:5), NH3 is actually the limiting reactant, contrary to the initial assumption in the question. Therefore, there will be no excess NH3. Instead, there will be excess O2. To find out how much O2 remains, calculate the moles of O2 that would have been required to react with 0.117 mol NH3:
- 4 mol NH3 needs 5 mol O2: 5 mol O2 / 4 mol NH3 = 1.25 mol O2/mol NH3
- 0.117 mol NH3 would need 0.117 mol NH3 × 1.25 mol O2/mol NH3 = 0.146 mol O2.
Since only 0.125 mol O2 is available, all of it will react. Thus, there is no O2 left over, and there's no NH3 left over because it's the limiting reactant. So, no excess reactant remains after the reaction has stopped.