Final answer:
To produce 45.0 kg of ammonia gas with a 74.0% yield, approximately 178.9 kg of nitrogen gas is required, based on stoichiometry and conversion between moles and grams.
Step-by-step explanation:
To calculate the amount of nitrogen gas (N2) needed to produce 45.0 kg of ammonia gas (NH3) with a 74.0% yield, we first need to understand the stoichiometry of the balanced chemical equation for the production of ammonia through the Haber process:
N2 + 3 H2 → 2 NH3
This equation indicates that 1 mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia gas. Based on the molar masses of these gases (N2 has a molar mass of 28 g/mol and NH3 has a molar mass of 17 g/mol), we can calculate the theoretical amount of NH3 produced per mole of N2 and then find out how much N2 would be needed for 45.0 kg of NH3.
To find the theoretical weight of NH3 produced by 1 mole of N2, we calculate: (2 moles NH3/1 mole N2) × (17 g NH3/mole NH3) = 34 g NH3.
Given the 74.0% yield, we divide the target production weight of NH3 by this yield percentage to obtain the theoretical production amount: 45.0 kg NH3 / 0.74 = 60.81 kg NH3 (theoretical).
To find the amount of N2 required, we invert the previous ratio to calculate the weight of N2 needed to produce the theoretical weight of NH3: (1 mole N2/34 g NH3) × (60.81 kg NH3 × 1000 g/kg) = 178.847 kg N2.
Therefore, it would take approximately 178.9 kg of nitrogen gas to produce 45.0 kg of ammonia gas with a 74.0% yield.