Question

Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor: 40.0 g NH₃ and 50.0 g O₂ are allowed to react according to the following equation:

4NH₃ + 5O₂ à 4NO + 6H₂O

4. What is the maximum number of grams of nitric oxide that can be produced in this experiment?

                     a. 50.0 g

                     b. 70.6 g                       

                     c. 37.5 g

                     d. 40.0 g

                     e. 58.6 g

Answer 1

Solution :-

Balanced reaction equation

4NH3 + 5O2----- > 4NO+6H2O

40.0 gNH3 and 50.0 g O2.

Maximum mass of NO produced = ?

Lets first calculate moles of the NH3 and O2

Moles = mass / molar mass

Moles of NH3 = 40.0g / 17.03 g per mol

= 2.349 mol NH3

Moles of O2 = 50.0 g / 32 g permol

= 1.5625 mol O2

Now lets calculate the moles of the O2 needed to react with the 2.349 mol NH3

Mole ratio of the NH3 to O2 is 4 :5

2.349mol NH3 * 5 mol O2 / 4 mol NH3 = 2.936 mol O2

moles of the O2 needed are more than moles of O2 available therefore the O2 is the limiting reactant

now lets calculate the moles of the NO using the moles of the O2

mole ratio of the O2 to NO is 5 :4

1.5625 mol O2 * 4 mol NO / 5 mol O2 = 1.25 mol NO

Now lets convert moles of NO to its mass

Mass = moles * molar mass

Mass of NO = 1.25 mol NO * 30.01 g per mol

                      = 37.5 g NO

Therefore the correct answer is option C that is maximum amount of the NO that can be produced by the reaction is 37.5 g