Answer:47.05% is the percent yield of nitrogen in the reaction.
Step-by-step explanation:
heoretical yield of nitrogen gas = x
Moles of ammonia =
According to reaction,2 moles of ammonia gives 1 mol of nitrogen gas.
Then 2.3529 mol of ammonia will give:
of nitrogen gas
Mass of 1.1764 moles of nitrogen gas,x = 1.1764 mol × 28 g/mol=32.94 g
Experiential yield of nitrogen gas = 15.5 g
Percentage yield:
hope that help
47.05% is the percent yield of nitrogen in the reaction.