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Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor:
4NH3 + 502 ---> 4NO + 6H2O
What is the theoretical yield of water, in moles, when 40.0 g NH3 and 50.0g, O2 are mixed and allowed to react?
Molar mass of NH3,
MM = 1*MM(N) + 3*MM(H)
= 1*14.01 + 3*1.008
= 17.034 g/mol
mass(NH3)= 40.0 g
use:
number of mol of NH3,
n = mass of NH3/molar mass of NH3
=(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 = mass of O2/molar mass of O2
=(50 g)/(32 g/mol)
= 1.562 mol
Balanced chemical equation is:
4 NH3 + 5 O2 ---> 6 H2O + 4 NO
4 mol of NH3 reacts with 5 mol of O2
for 2.348 mol of NH3, 2.935 mol of O2 is required
But we have 1.562 mol of O2
so, O2 is limiting reagent
we will use O2 in further calculation
According to balanced equation
mol of H2O formed = (6/5)* moles of O2
= (6/5)*1.562
= 1.875 mol
Answer: 1.87 mol
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