In a lab, nitrogen trichloride is synthesized from its elements. This reaction has a 60.0% yield...

Question

In a lab, nitrogen trichloride is synthesized from its elements. This reaction has a 60.0% yield. A chemist wants to prepare 50.0g of nitrogen trichloride, with a 60.0% yield, what mass of nitrogen should be used for the reaction?

The answer is 9.70g N₂ according to the answer sheet, but I'm not sure how to get there.

Answer 1

Hello, Anoymoose,

We need to start with a balanced equation:

1N₂ + 3Cl₂ = 2NCl₃

Now let's convert the 50.0 grams of NCl₃ to moles by dividing by the molar mass of NCl₃ (120.4g/mole). I get 0.415 moles NCl₃. Since we only get 60% yield, we need to dividie the moles NCl₃ by 0.60, to account for the loss. That means we need to plan to make 0.692 moles of NCl₃, undoubtedly due to the clumsiness of the lab technician. The balanced equation tells us we need 1 mole of N₂ for every 2 moles of NCl₃ (a mole ratio of 1/2). So we need 1/2 of 0.692 moles, or 0.346 moles of N₂. Multiply by nitrogen's molar mass of 28 (N₂) to find grams N₂ required. That looks VERY close to 9.70 grams. (It is).

Bob