Question

Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia.

N2(g)+3H2(g)⟶2NH3(g)
Assume 0.230 mol N2 and 0.758 mol H2 are present initially.PLEASE SHOW steps!!

1)After complete reaction, how many moles of ammonia NH3 are produced?

2)How many moles of H2 remain?

3)How many moles of N2 remain?

4)What is the limiting reactant?
nitrogen or
hydrogen

Answer 1

1)

Balanced chemical equation is:

N2 + 3 H2 ---> 2 NH3 +

1 mol of N2 reacts with 3 mol of H2

for 0.23 mol of N2, 0.69 mol of H2 is required

But we have 0.758 mol of H2

so, N2 is limiting reagent

we will use N2 in further calculation

According to balanced equation

mol of NH3 formed = (2/1)* moles of N2

= (2/1)*0.23

= 0.46 mol

Answer: 0.460 mol

2)

According to balanced equation

mol of H2 reacted = (3/1)* moles of N2

= (3/1)*0.23

= 0.69 mol

mol of H2 remaining = mol initially present - mol reacted

mol of H2 remaining = 0.758 - 0.69

mol of H2 remaining = 6.8*10^-2 mol

Answer: 0.0680 mol

3)

N2 is limiting reagent

So, no N2 is present

Answer: 0.0 mol

4)

N2 is limiting reagent

Answer: Nitrogen