Question

NH3 reacts w/ F2 to produce solid N2F4 and HF. A reaction mixture contains 4.00 g NH3 and 14.0 g F2.
A. Which reactant is limiting?
B. What is the theoretical yield of NaH from the above reaction mixture?
C. What is the percent yield if 4 g NaH is formed?

Answer 1

first, balance the equation.
NH3 + F2 → N2F4 + HF
clearly, you need 2N on the left, so
2NH3 + F2 → N2F4 + HF
Now you have 6H on the left, so
2NH3 + F2 → N2F4 + 6HF
Finally, you have 2F on the left, and 10F on the right, so
2NH3 + 5F2 → N2F4 + 6HF
(A) Now determine the number of moles of the reactants in the given masses; if you don't have at least twice as many moles of NH3 as you do F2, then NH3 will be the limiting reactant.

and now finish it off. Post your work if you get stuck.