Question

Question 14

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, 2.10 g of NH3 reacts with 3.85 g of O2.

Part B

How many grams of NO and of H2O form?

Enter your answers numerically separated by a comma.

mNO,mH2O =

g

Part C

How many grams of the excess reactant remain after the limiting reactant is completely consumed?

m =

Answer 1

Let is first calculate the number of moles of NH3 and moles of O2 used in the reaction,

The molar mass of NH3 and O2 are 15 g/mol and 32 g/mol, respectively

The number of moles of NH3 = Weight of NH3 / molar mass NH3

                                                  = 2.10 g / 15 g/mol = 0.14 moles

The number of moles of O2 = Weight of O2 / molar mass of O2

       = 3.85 g / 32 g/mol = 0.12 mol

From the balanced chemical equation it is clear that 4 mole of NH3 requires 5 moles of O2

Therefore, the ideal ratio of number of moles of NH3 and O2 = 4/5 = 0.8

The actual ratio of number of moles of NH3 and O2 = 0.14 / 0.12 = 1.17

As the actual ratio of number of moles of CaCO3 and O2 is greater than the ideal ratio, NH3 is used in excess and O2 is the used less in the reaction which is a limiting reagent. The mass of NO and H2O formed will depend on the number of moles of O2 used.

From the balanced chemical equation it is clear that 5 moles of O2 produces 4 mole of NO. Therefore, we can consider,

5 moles of O2   = 4 mole of NO

For, 0.12 moles of O2 = (0.12 x 4)/5 = 0.096 moles of NO

The mass of NO formed = Number of moles of NO formed x molecular weight of NO in g/mol

The molecular weight of NO is 30 g/mol

The mass of NO formed = 0.096 mol x 30 g/mol

= 2.88 g

Therefore 2.88 g of NO will be formed.

From the balanced chemical equation it is clear that 5 moles of O2 produces 6 mole of H2O. Therefore, we can consider,

5 moles of O2   = 6 mole of H2O

For, 0.12 moles of O2 = (0.12 x 6)/5 = 0.144 moles of H2O

The mass of H2O formed = Number of moles of H2O formed x molecular weight of H2O in g/mol

The molecular weight of H2O is 18 g/mol

The mass of NO formed = 0.144 mol x 18 g/mol

= 2.6 g

Therefore 2.6 g of H2O will be formed.

As O2l is getting consumed completely in the reaction, the excess reagent is NH3.

The excess mass of NH3 = Mass of NH3 used in g - Mass of NH3 consumed in g

The mass of NH3 consumed in g can be calculated from number of moles of O2 used. As, from the balanced chemical equation 5 moles of O2 consumes 4 mole of NH3.

Therefore, we can consider,

5 moles of O2 used = 4 mole of NH3 consumed

For, 0.12 moles of O2 = (0.12 x 4) / 5 = 0.0.96 moles of NH3 consumed

Mass of NH3 consumed = number of moles of NH3 consumed x molecular weight of NH3 in g/mol

= 0.0.96 mol x 15 g/mol

= 1.44 g

The excess mass of NH3 = 2.1 g – 1.44 g = 0.66 g

Therefore, 0.66 g of excess reactant will remain after completion of reaction