when 40. 0 g of ammonia and 50. 0 g of oxygen are mixed and allowed to react .1.57 moles is the theoretical yield of water and 1.87 moles in balanced chemical
= 1*14.01 + 3*1.008
= 17.034 g/mol
mass(NH3)= 40.0 g
use:
number of mol of NH3,
n = molar mass of ammonia
=(40 g)/(17.03 g/mol)
= 2.348 mol
Molar mass of O2 = 32 g/mol
mass(O2)= 50.0 g
use:
number of mol of O2,
n = molar mass of oxygen
=(50 g)/(32 g/mol)
1.562 mol or 1.57 mol
Balanced chemical equation is:
4 NH₃ + 5 O₂ -> 6 H₂O + 4NO
4 mol of NH3 reacts with 5 mol of oxygen for 2.348 mol of NH3, 2.935 mol of O2 is required But we have 1.562 mol of O2 so, use O2 in next calculation
According to balanced equation mol of H2O formed = (6/5)* moles of oxygen
= (6/5)*1.562
= 1.875 mol
= 1.87 mol
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