Question

Mass of H2O possible from NH3 when the balanced equation 4NH3 + 5O2 yields 4NO + 6H2O

Answer the following based on beginning the reaction with 1.92 g NH₃ and 8.95 g O₂ and recovering 2.32 g of H₂O at the end of the reaction.

Answer 1

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

number of moles of NH3 = 1.92g / 17.031 g.mol^-1 = 0.112 mole

from the balanced equation we can say that

4 mole of NH3 produces 6 mole of H2O so

0.112 mole of NH3 will produce

= 0.112 mole of NH3*(6 mole of H2O/4 mole of NH3)

= 0.168 mole of H2O

1 mole of H2O = 18.015g

0.168 mole of H2O = 3.03 g

Therefore, the mass of H2O = 3.03 g

number of moles of O2 = 8.95g / 32.0 g.mol^-1 = 0.280 mole

from the balanced equation we can say that

4 mole of NH3 requres 5 mole of O2 so

0.112 mole of NH3 will require 0.14 mole of O2

we have 0.280 mole of O2 so O2 is excess reactant

number of moles of excess reactant = 0.280 - 0.14 = 0.14 mole of O2

1 mole of O2 = 32.0 g

so 0.14 mole of O2 = 4.48g

Therefore, the mass of excess reactant = 4.48g