If 8.00 moles of NH₃ of and 10.00 moles of O₂ react in the following reaction, how many moles of which reactant will be left over? 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g) A) 2.00 mol O₂
B) 2.00 mol NH₃
C) 8.00 mol NO
D) no moles will be left over of either reactant because both are limiting.

Option (D) no moles will be left over of either reactant because both are limiting

Explanation:

The balanced equation for the reaction is given below:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

From the balanced equation above,

4 moles of NH₃ reacted with 5 moles of O₂.

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

From the balanced equation above,

4 moles of NH₃ reacted with 5 moles of O₂.

Therefore, 8 moles of NH₃ will react with = (8 × 5)/4 = 10 moles of O₂.

From the calculations made above, we see clearly that 10 moles of O₂ reacted completely with 8 moles of NH₃. This implies that both reactant are limiting reactant and there is no excess reactant.

Thus, no reactant is left over.

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