Question

determine the mass of water formed when 21.5 L NH3 (at 298 K and 1.50 atm) is reacted with 38.9 L of O2 (at 323 K and 1.1 atm).
4NH3 + 5O2=4NO+6H2O

Answer 1

PV = nRT

n = PV/RT

= 1.5*21.5/0.0821*298   = 1.32moles of NH3

PV = nRT

n = PV/RT

    = 1.1*38.9/0.0821*323 = 1.6moles

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

4 moles of NH3 react with 5 moles of O2

1.32 moles of NH3 react with = 5*1.32/4 = 1.65 moles of O2

O2 is limiting reagent

5 moles of O2 react with NH3 to gives 6 moles of H2O

1.6moles of O2 react with NH3 to gives = 1.6*6/5 = 1.92 moles of H2O

mass of H2O = no of moles * gram molar mass

                     = 1.92*18 = 34.56g of H2O