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3.6 moles

Explanation:

The balanced equation for the reaction is given below:

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

1 mole of N₂ reacted with 3 moles of H₂ to produce 2 moles NH₃.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

1 mole of N₂ reacted with 3 moles of H₂.

Therefore, 3.2 moles of N₂ will react with = 3.2 × 3 = 9.6 moles of H₂.

From the calculation made above, we can see that it will take a higher amount (i.e 9.6 moles) of H₂ than what was given (i.e 5.4 moles) to react completely with 3.2 moles of N₂.

Therefore, H₂ is the limiting reactant and N₂ is the excess reactant.

Finally, we the greatest quantity of ammonia, NH₃ produced from the reaction.

In this case, the limiting reactant will be use because all of it is consumed in the reaction.

The limiting reactant is H₂ and greatest quantity of ammonia, NH₃ produced can be obtained as follow:

From the balanced equation above,

3 moles of H₂ reacted to produce 2 moles NH₃.

Therefore, 5.4 of H₂ will react to produce = (5.4 × 2)/3 = 3.6 moles of NH₃

Thus, the greatest quantity of ammonia, NH₃ produced from the reaction is 3.6 moles

14,16 18 reacts with group