28 g of N2 and 6g of H2 were mixed. At equilibrium, 17g NH3 w

1.	28 g of N2 and 6g of H2 were mixed. At equilibrium, 17g NH3 was produced. The weights of N2 and H2 at equilibrium are respectively.
Answer» The reaction for the formation of ammonia is N₂+3H₂⇔ 2NH₃. 1 mole (28 g) of nitrogen will react with 3 moles (6g) of hydrogen to form 2 moles (34g) of ammonia if the reaction is 100% completed. But only 17 g of ammonia are actually formed. Hence, the reaction is 50% complete. Hence, 0.5 mole (14 g) of nitrogen and 3 moles (3g) of hydrogen will remain unreacted at equilibrium.

28 g of N2 and 6g of H2 were mixed. At equilibrium, 17g NH3 was produced. The weights of N2 and H2 at equilibrium are respectively.

Answer»

The reaction for the formation of ammonia is N₂+3H₂⇔ 2NH₃.

1 mole (28 g) of nitrogen will react with 3 moles (6g) of hydrogen to form 2 moles (34g) of ammonia if the reaction is 100% completed.

But only 17 g of ammonia are actually formed.

Hence, the reaction is 50% complete.

Hence, 0.5 mole (14 g) of nitrogen and 3 moles (3g) of hydrogen will remain unreacted at equilibrium.

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28 g of N2 and 6g of H2 were mixed. At equilibrium, 17g NH3 was produced. The weights of N2 and H2 at equilibrium are respectively.

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