The answer to your question is a 12
Ignoring the n's, there would only be one unpaired electron.
Answer:
Cl₂ is the limiting reactant.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
N₂ + 3Cl₂ —> 2NCl₃
From the balanced equation above,
1 L of N₂ reacted with 3 L of Cl₂.
Finally, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
1 L of N₂ reacted with 3 L of Cl₂.
Therefore, 134 L of N₂ will react with = 134 × 3 = 402 L of Cl₂.
From the calculation made above, we can see that a higher volume (i.e 402 L) of Cl₂ than what was given (i.e 99 L) is needed to react completely with 134 L of N₂.
Therefore, Cl₂ is the limiting reactant and N₂ is the excess reactant.
Answer:
Add copper (II) oxide (insoluble base), a little at a time to the warm dilute sulfuric acid and stir until the copper (II) oxide is in excess (stops disappearing) Filter the mixture into an evaporating basin to remove the excess copper (II) oxide. Leave the filtrate in a warm place to dry and crystallize.
