Final answer:
To find how much nitrogen monoxide is produced, we first calculate the moles of O2 and NH3, determine the limiting reactant, and use the mole ratio to calculate the mass of NO. In this case, we find that O2 is the limiting reactant and the calculated production of NO is 112.54 g, which is not matched by any of the given answer choices.
Step-by-step explanation:
To determine how much nitrogen monoxide (NO) will be produced from 150.0 g of O2 and 75.0 g of NH3, we must first calculate the moles of each reactant, and then use the mole ratio from the balanced chemical equation 4 NH3 + 5 O2 -> 4 NO + 6 H2O to find the limiting reactant.
Molar masses (g/mol): O2 = 32.00, NH3 = 17.03, NO = 30.01
Moles of O2: 150.0 g / 32.00 g/mol = 4.6875 mol
Moles of NH3: 75.0 g / 17.03 g/mol = 4.4046 mol
Using the balanced equation, 5 mol of O2 reacts with 4 mol of NH3. Therefore, we can calculate the amount of NH3 needed to react with 4.6875 mol O2:
4.6875 mol O2 * (4 mol NH3 / 5 mol O2) = 3.75 mol NH3 (needed)
Since we only have 4.4046 mol NH3 and need 3.75 mol to react with the available O2, NH3 is in excess. Hence, O2 is the limiting reactant. Now we determine the amount of NO produced by using the mole ratio between O2 and NO:
4.6875 mol O2 * (4 mol NO / 5 mol O2) = 3.75 mol NO
Then convert moles of NO to grams:
3.75 mol NO * 30.01 g/mol NO = 112.54 g NO
None of the given answer choices match this result. If these are the only answer choices provided in a multiple-choice setting, it could indicate the presence of a typo or error in the question or answers or a misunderstanding in the calculation process.