Hello, Anoymoose,

We need to start with a balanced equation:

1N_{2} + 3Cl_{2} = 2NCl_{3}

Now let's convert the 50.0 grams of NCl_{3} to moles by dividing by the molar mass of NCl_{3} (120.4g/mole). I get 0.415 moles NCl_{3}. Since we only get 60% yield, we need to dividie the moles NCl_{3} by 0.60, to account for the loss. That means we need to plan to make 0.692 moles of NCl_{3}, undoubtedly due to the clumsiness of the lab technician. The balanced equation tells us we need 1 mole of N_{2} for every 2 moles of NCl_{3} (a mole ratio of 1/2). So we need 1/2 of 0.692 moles, or 0.346 moles of N_{2}. Multiply by nitrogen's molar mass of 28 (N_{2}) to find grams N_{2} required. That looks VERY close to 9.70 grams. (It is).

Bob