Question

7g of N2 is allowed to react with 2g of H2 to form NH3.identify the limiting reagent? calculate the amount of NH3 formed​

Answer 1

The balanced checmical equation, for the reaction is :-

N₂ + 3H₂ → 2NH₃

Number of mole of N₂ :-

= Given Mass/Molar mass

= 7/28

= 0.25 mole

Number of mole of H₂ :-

= Given Mass/Molar mass

= 2/2

= 1 mole

Here, we are given more moles of H₂ than of N₂ therefore, H₂ is in excess and some of it will remain unreacted when the reaction is over.

Hence, N₂ is the limiting reagent and will control the amount of product.

Now, from the equation :-

∵ 28g N₂ → 34g NH₃

∴ 7g N₂ → 34×7/28 = 8.5g NH₃

Thus, 8.5g of NH₃ is formed.

Answer 2

Here, we are given more moles of H₂ than of N₂ therefore, H₂ is in excess and some of it will remain unreacted when the reaction is over. Hence, N₂ is the limiting reagent and will control the amount of product.