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, 1.60 g of NH3 reacts with 2.93 g of O2.

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

Answer 1

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, 1.60 g of NH3 reacts with 2.93 g of O2.

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

We find limiting reactant by using moles of each reactant

Mol NH3 = Mass in g / Molar mass = 1.60 g / 17.0307 g per mol = 0.09395 Mol NH3

Mol O2 = 2.93 g/ 31.998 g per mol = 0.9157 mol O2

We calculate moles of one of the product by using both reactant

Lets find moles of NO

1. By using moles of NH3

Mole ratio of NH3 : NO is 4 : 4 or 1:1

Moles of NO = moles of NH3 * 1mol NO / 1mol NH3

= 0.09395 mol NH3 * 1mol NO / 1 mol NH3

= 0.09395 mol NO

1. Mol of NO from mol of O2

Mol ratio of O2 : NO is 5 : 4

= 0.9157 mol O2 * 4 mol NO / 5 mol O2

= 0.0733 mol

O2 gives less number of moles of NO so O2 is limiting reactant

NH3 is excess reagent and we find moles of NH3 required to react with 0.09157 mol of O2

Mol of NH3 reacted

Mol ratio of NH3 to O2 is 4 : 5

Mol of NH3 reacted = mol of O2 * 4 mol NH3 / 5 mol O2

= 0.09157 mol * 4 mol NH3 / 5 mol O2

= 0.0733 mol NH3

Mols of NH3 remained = 0.09395 mol – 0.0733 mol = 0.020693 mol NH3

Mass of NH3 in g

= mol of NH3 * molar mass of NH3

= 0.020693 mol * 17.0307 g per mol

= 0.3524 g

Amount of NH3 remained = 0.3524 g