Question

In an experiment, 6.25 g of NH₃ are allowed to react with 8.50 g of O₂. The chemical equation for the reaction is 4NH₃ + 5O₂ → 4NO + 6 H₂O. How many grams of water are produced? Which reactant limits the production of water? Which reactant is in excess?

Answer 1

   4NH3 + 5O2 → 4NO + 6 H2O

Molar mass(g/mol) 17 32 30 18

According to the balanced equation ,

4mol=4x17 g of NH3 reacts with 5mol=5x32 g of O2

M g of NH3 reacts with 8.50 g of O2

M = (8.50x4x17)/(5x32)

= 3.61 g

So 6.25-3.61 = 2.64 g of NH3 left unreacted so it is excess reactant

Since all the mass of O2 completly reacted it is limiting reactant.

Again from the balanced equation ,

5 mol = 5x32 g of O2 produces 6 mol = 6x18 g of water

8.50 g of O2 produces N g of water

N = ( 8.50x6x18) / (5x32)

= 5.74 g of water