Chemistry My Lab

Question

3H2(g)+N2(g)-> 2NH3(g)

1.26 g H2 is allowed to react with 10.3g N2, producing 2.56 g NH3.

Part A- What is theoretical yield in grams for this reaction under the given conditions ?

Part B- what is the percent yield for this reaction under the given condition ?

Answer 1

Hello, Grace,

The masses for H₂ and N₂ can be converted into moles of each by dividing by their respective molar masses (2 and 28 g/mole, respectivelly).

H₂ 0.81 mole

N₂ 0.471 mole

The equation tells us we need 3 moles of H₂ for every 1 mole of N₂. There will not be enough hydrogen to completely react with the nitrogen, so it is the limiting reagent. Let's assume all the H₂ is consumed (0.81 mole). The molar ratio of N₂/H₂ is (1/3), se we will consume (1/3)*(0.471 mole of N₂), or 0.27 mole N₂. That will leave (0.471 - 0.27) or 0.201 moles of unreacted nitrogen. (multiply by nitrogen's molar mass of 28 g/mole to determine the leftowers in grams).

Using moles H₂ as the limiting reagent, we can find the amount of NH₃ we'd expect. The molar ratio of NH₃ to H₂ is (2/3). The equation says we'll get 2 moles NH₃ for every 3 moles H₂. Take the moles H₂ consumed as we did above, and multiply it by this ratio:

(0.81 mole H₂)*((2 moles NH₃)/ (3 mole H₂). = 0.54 moles NH₃

This is equal to (0.54 moles NH₃)*(17 g NH₃/mole NH₃) = 9.18 grams, the theorectical yield.

We only got 2.56 grams of NH₃, so the actual yield is only (2.56/9.18) or 27.8%

That stinks, so maybe I did a calculation incorrectly. But the process should be OK. Please let me know if you get the same number. In any event, NH₃ does stink, so a low yield might be appreciated by others around you.

Bob