If 25.0 g of NH₃ and 45.0g of O₂ react in the following reaction, what is the mass in grams of NO that will be formed? 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g)

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

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