From the above reaction 1 mol of N2(g) gives 2 mol 0f NH3(g). So 3 mol of N2(g) should give 6 mol or (6x17) 102 g of NH3(g). But we are getting only 85.0 g of NH3.
Hence %yield of NH3 = 100x85.0/102 = 83.3%
Answer D. is correct.
2. 2HgO (s) --------------------- 2Hg (l) + O2 (g)
2 mol 2 mol 1 mol
4 mol 4 mol 2 mol
As only 1.5 mol of O2 is obtained, so % yield = 1.5x100/2 = 75.0%
Moreover 603 g of Hg = 603/200.59 = 3.00 molof Hg. Considering this also
This is the chemical reaction for above experiment.
A mixture of nitrogen gas and hydrogen gas reacts in a closed container to form ammonia, . The reaction ceases before either reactant has been totally consumed.
Molybdenum as promoter too can enhance the effeciency of the catalyst.
