Question

Consider the following balanced reaction between hydrogen and nitrogen to form ammonia: 3H2(g) + N2(g)→2NH3(g)
How many moles of NH3 can be produced from 24.0 mol of H2 and excess N2?

Answer 1

the balanced equation for the formation of ammonia is as follows

3H₂ + N₂ --> 2NH₃

we are told that N₂ is in excess. this means that H₂ is the limiting reactant

limiting reactant is fully consumed in the reaction , amount of product formed depends on amount of limiting reactant present

stoichiometry of H₂ to NH₃ is 3:2

when 3 mol of H₂ reacts - 2 mol of NH₃ are formed

therefore when 24.0 mol of H₂ reacts - 2/3 x 24.0 = 16 mol of NH₃ are formed

16 mol of NH₃ are formed

Answer 2

The number of moles of NH₃ that can be produced from 24.0 mol of H₂ and excess N₂ is 16 moles.

What is Limiting Reagent ?

The limiting reagent in a chemical reaction is a reactant that is totally consumed when the chemical reaction is completed.

The amount of product formed is limited by this reagent, since the reaction cannot continue without it.

The balanced equation for the reaction is ;

3H₂(g) + N₂(g) --> 2NH₃(g)

The equation shows when 3 moles of hydrogen reacts 2 moles of ammonia is formed.

3 moles of H₂……….2 moles of NH₃

So when there is 24 moles of hydrogen, no. of moles of NH₃ will be ;

NH₃ = 24 x 2/3 = 8 x 2 = 16 moles

Therefore, The number of moles of NH₃ that can be produced from 24.0 mol of H₂ and excess N₂ is 16 moles.

Learn more about Limiting reagent here :

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