Answer:
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
