Question

If 85.0 g sodium is mixed with 29.0 g nitrogen gas, what mass of sodium nitride forms?

If the reaction in part B has a percent yield of 73.0 %, how much sodium nitride is actually produced?

Answer 1

\[6Na + N _{2} \rightarrow 2Na _{3}N\]

First, determine which reactant is the limiting reactant by using stoichiometry to determine the amount of moles of the product produced by the amount of each reactant given (Making sure to remember that elemental nitrogen is diatomic):


29 g Nitrogen 1 mole Nitrogen 2 moles Sodium Nitride
------------ x -------------- x -------------------- = 1.04 moles
1 28.014 g 2 moles Nitrogen

85 g Sodium 1 mole Sodium 2 moles Sodium Nitride
----------- x ------------- x --------------------- = 1.23 moles
1 22.990 g 6 moles Nitrogen


Since the 29 g of nitrogen produces less sodium nitride, it is the limiting reactant, and 1.04 moles of sodium nitride are produced.

To determine the actual yield, multiply the theoretical yield by the percent yield:

1.04 moles x (73/100) = 0.756 g Sodium Nitride