Answer:
The correct answer would be : 33.8 g
Explanation:
Molar mass of ammonia,
Molar Mass = 1* Molar Mass(N) + 3* Molar Mass (H)
= 1*14.01 + 3*1.008 = 17.034 g/mol
mass(NH3)= 25.0 g (given)
number of mol of NH3,
n = mass of NH3/molar mass of NH3
=(25.0 g)/(17.034 g/mol)
= 1.468 mol
Now,
Molar mass of O2
= 32 g/mol
mass(O2)= 45.0 g
similar as ammonia
n (O2)=(45.0 g)/(32 g/mol)
= 1.406 mol
Balanced chemical equation is:
4 NH3 + 5 O2 ---> 4 NO + 6 H2O
1.83456 mol of O2 is required for 1.46765 mol of NH3
by the calculation we have only 1.40625 mol of O2
Thus, the limiting agent will be - O2
now the Molar mass of NO,
= 1*14.01 + 1*16.0
= 30.01 g/mol (similar formula used for NH3)
Balanced equation
:
mol of NO formed = (4/5)* moles of O2
= (4/5)×1.40625 (from above calculation)
= 1.125 mol
mass of NO = number of moles × molar mass
= 1.125*30.01
= 33.8 g
Thus, the correct answer would be : 33.8 g
