Answer:
Molar mass of NH3,
MM = 1*MM(N) + 3*MM(H)
= 1*14.01 + 3*1.008
= 17.034 g/mol
mass(NH3)= 25.0 g
use:
number of mol of NH3,
n = mass of NH3/molar mass of NH3
=(25 g)/(17.03 g/mol)
= 1.468 mol
Molar mass of O2 = 32 g/mol
mass(O2)= 35.8 g
use:
number of mol of O2,
n = mass of O2/molar mass of O2
=(35.8 g)/(32 g/mol)
= 1.119 mol
The balanced chemical equation is:
4 NH3 + 5 O2 ---> 4 NO + 6 H2O
4 mol of NH3 reacts with 5 mol of O2 for 1.468 mol of NH3, 1.835 mol of O2 is required
But we have 1.119 mol of O2
so, O2 is limiting reagent
we will use O2 in further calculation
Molar mass of NO,
MM = 1*MM(N) + 1*MM(O)
= 1*14.01 + 1*16.0
= 30.01 g/mol
According to balanced equation
mol of NO formed = (4/5)* moles of O2
= (4/5)*1.119 = 0.895 mol
use: mass of NO = number of mol * molar mass
= 0.895*30.01
= 26.86 g
Answer: 26.9 g
