Question

What mass of NH₃ is required to produce 125 g of NO₂ if the percentage yield for the reaction is 67.2%? Enter your answer in grams and without units

. 4 NH₃(g) + 7 O₂(g) → 4 NO₂(g) + 6 H₂O(l)

Answer 1

To produce 125 g of NO₂ with a 67.2% yield, 68.8 g of NH₃ is required.

Step-by-step explanation:

The question is asking how much mass of NH₃ (ammonia) is needed to produce 125 g of NO₂ (nitrogen dioxide) with a percent yield of 67.2%. To find out, we first need to use the molar mass of NO₂ to convert grams to moles, as the stoichiometry of the reaction operates on a molar basis. The molar mass of NO₂ is approximately 46.0055 g/mol, so 125 g corresponds to:

125 g NO₂ * (1 mol NO₂ / 46.0055 g NO₂) = 2.717 moles NO₂

According to the balanced chemical equation, the molar ratio of NH₃ to NO₂ is 1:1. So, 2.717 moles of NH₃ are needed theoretically to produce 2.717 moles of NO₂. The molar mass of NH₃ is approximately 17.031 g/mol, so the mass of NH₃ needed theoretically is:

2.717 moles NH₃ * 17.031 g NH₃/mol = 46.27 g NH₃

However, since the percent yield is 67.2%, the actual amount of NH₃ required is:

46.27 g NH₃ / 0.672 = 68.8 g NH₃ (to two decimal places)