In the reaction,
2NH3+F2⟶N2F4+6HF
16.85g of N2F4 is obtained by mixing 2gNH3 and 8g F2. The percentage yield of the production is:
77%
F2 is limiting reagent.
38g of flourine produces 104g of N2H4
8g of flourine produces 8×10438=21.89g of N2H4
%yield = 16.8521.89×100 =77%