Question I or IU
2 answers:
You might be interested in
A. The limiting reagent in the reaction is N₂
B. The number of mole of NH₃ molecules produced is 4 moles
C. The number of H₂ molecules in excess is 1 mole
<h3>A. How to determine the limiting reactant </h3>From the diagram,
- Mole of N₂ = 2 moles
- Mole of H₂ = 7 moles
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
1 mole of N₂ requires 3 moles of H₂.
Therefore,
2 moles of N₂ will require = 2 × 3 = 6 moles of H₂
From the above illustration, we can see that only 6 moles out of 7 moles of H₂ given is required to react completely with 2 moles of N₂.
Therefore, N₂ is the limiting reactant
<h3>B. How to determine the mole of NH₃ produced </h3>In this case, the limiting reactant will be used. This is illustrated below:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
1 mole of N₂ reacted to produce 2 moles of NH₃.
Therefore,
2 moles of N₂ will react to produce = 2 × 2 = 4 moles of NH₃.
Thus, 4 moles of NH₃ molecules were obtained from the reaction.
C. How do determine the excess reactant
From A above, we discovered that N₂ is the limiting reactant.
Therefore, H₂ will be the excess reactant.
<h3>Further proof</h3>- Mole of H₂ given = 7 moles
- Mole of H₂ that reacted = 6 moles
- Excess of H₂ =?
Excess of H₂ = 7 – 6
Excess of H₂ = 1 mole
<h3>Complete question:</h3>See attached photo
Learn more about stoichiometry:
