Question

Be sure to answer all parts.

Ammonia is produced by the reaction of nitrogen and hydrogen according to the equation

N₂(g) + 3H₂(g) → 2NH₃(g)

1. Calculate the mass of ammonia produced when 30.0 g of nitrogen react with 11.3 g of hydrogen.

2. Which is the excess reactant and how much of it will be left over when the reaction is complete?

Answer 1

N₂(g) + 3H₂(g) → 2NH₃(g)

no of moles of N2   = W/G.M.Wt

                                 = 30/28   = 1.07 moles

no of molesof H2    = W/G.M.Wt

                                = 11.3/2   = 5.65 moles

3 moles of H2 react with 1 mole of N2

5.65 moles of H2 react with = 1*5.65/3   = 1.883 moles of N2 is required

N2 is limiting reactant

N₂(g) + 3H₂(g) → 2NH₃(g)

1 mole of N2 react with excess of H2 to gives 2 moles of NH3

1.07 moles of N2 react with excess of h2 to gives = 2*1.07/1   = 2.14 moles of NH3

mass of NH3 = no of moles * gram molar mass

                        = 2.14*17    = 36.38g of NH3 >>>>naser

2.

N₂(g) + 3H₂(g) → 2NH₃(g)

1 mole of N2 react with 3 moles of H2

1.07 moles of N2 react with = 3*1.07/1   = 3.21 moles of H2

H2 is excess reactant

The no of moles of exces reactant left over after complete the reaction = 5.65-3.21 = 2.44 moles

The amount of exces reactant left over after complete th reaction = 2.44*2 = 4.88g of H2