Question

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of NH3 to NO: 4NH3(g)+5O2(g)→4NO(g)+6H2O(g) In a certain experiment, 2.05 g of NH3 reacts with 3.76 g of O2.

How many grams of NO and of H2O form? Enter your answers numerically separated by a comma.

Answer 1

Number of moles of NH3 = 2.05 g / 17.031 g/mol = 0.120 mole

Number of moles of O2 = 3.76 g / 32.0 g/mol = 0.118 mole

From the balanced equation we can say that

4 mole of NH3 requires 5 mole of O2 so

0.120 mole of NH3 will require

= 0.120 mole of NH3 *(5 mole of O2 / 4 mole of NH3)

= 0.150 mole of O2

But we have 0.118 mole of O2 which is in short so O2 is limiting reactant

From the balanced equation we can say that

5 mole of O2 produces 4 mole of NO so

0.118 mole of O2 will produce

= 0.118 mole of O2 *(4 mole of NO / 5 mole of O2)

= 0.0944 mole of NO

mass of 1 mole of NO = 30.01 g so

the mass of 0.0944 mole of NO = 2.83 g

Therefore, the mass of NO produced would be 2.83 g

From the balanced equation we can say that

5 mole of O2 produces 6 mole of H2O so

0.118 mole of O2 will produce

= 0.118 mole of O2 *(6 mole of H2O / 5 mole of O2)

= 0.142 mole of H2O

mass of 1 mole of H2O = 18.016 g so

the mass of 0.142 mole of H2O = 2.56 g

Therefore, the mass of H2O produced would be 2.56 g

The masses of NO and H2O produced would be 2.83 , 2.56