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 2.00 g of NH3 reacts with 2.50 g of O2. Which is the limiting reactant?

Answer 1

Answer:

O2 is the limiting reactant.

Explanation:

Step 1: Data given

Mass of NH3 = 2.00 grams

Mass of O2 = 2.50 grams

Molar mass NH3 = 17.03 g/mol

Molar mass O2 = 32 g/mol

Step 2: The balanced equation

4NH3(g) +  5O2 (g) → 4NO(g) +  6H2O (g)

Step 3: calculate moles NH3

Moles NH3 = mass NH3 / molar mass NH3

Moles NH3 = 2.00 grams / 17.03 g/mol

Moles NH3 = 0.117 moles

Step 4: Calculate moles O2

Moles O2 = mass / molar mass O2

Moles O2 = 2.50 grams / 32 g/mol

Moles O2 = 0.0781 moles

Step 5: Calculate the limiting reactant

For 4 moles NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed. (0.0781 moles). NH3 is in excess. There will react 4/5 * 0.0781 moles =  0.0625 moles

There will remain 0.117 - 0.0625 = 0.0545 moles NH3

O2 is the limiting reactant.