Question

n2 +3h2--2nh3.assume 0.280 monofilament n2 and 0.886mol of h2 are present initially.
after completion how many moles of h2 remains
how many moles of ammonia are produced?
how many moles of n2 remain?
what is the limiting reactant?

Answer 1

N2 + 3H2 ------> 2NH3

0.28 0.886 0

1 mole of N2 reacts with 3 moles of H2 and produces 2 moles of Ammonia

0.28 moles of N2 reacts with 0.28*3 = 0.84 moles of H2 and produces 0.56 moles of NH3

Mole of ammonia produced = 0.56

Moles of N2 remain = 0

Moles of H2 remains = 0.886 - 0.84 = 0.046

Limiting reactant is N2

The reactant which is completely consumed in the reaction is called Limiting reactant.

Here N2 completely consumed in the reaction

So ; N2 is the limitng reactant