N2 + 3 H2 → 2 NH3

If 3.45 moles of nitrogen gas (N2) are added to 4.85 moles of hydrogen gas (H2), then:

a) How many moles of ammonia (NH3) can form? 3.23 moles

b) What is the limiting reactant? H2

c) What is the excess reactant, and how many moles will be left?

N₂ + 3 H₂ → 2 NH₃

1 mole 3 moles 2 moles

3.45 moles 10.35moles 6.90 moles

So 3.45 moles of nitrogen will require 10.35 moles of hydrogen

But hydrogen available is only 4.85 so this gas is the limiting reagent and product will be formed according to its mass.

a)

3 moles of hydrogen can form 2 moles of ammonia

4.85 moles of hydrogen will form 2 x 4.85 / 3 moles of ammonia

= 3.23 moles.

b) hydrogen is the limiting reagent.

c) Excess reactant is nitrogen

3 moles hydrogen need 1 mole of nitrogen

4.85 moles of hydrogen will need 4.85 / 3 moles of nitrogen

= 1.62 moles of nitrogen.

excess nitrogen = 3.45 - 1.62

= 1.83 moles.