Final answer:
KOH is the limiting reagent in the chemical reaction between MgCl₂ and KOH as only 6 moles of KOH are available for 6 moles of MgCl₂, which would require 12 moles of KOH to react completely. MgCl₂ is in excess and the reaction will stop when all KOH is consumed. The correct option is (d)
Step-by-step explanation:
In the reaction ℓgCl₂ + 2KOH → Mg(OH)₂ + 2KCl, the question is to identify the limiting reagent when 6 moles of MgCl₂ are added to 6 moles of KOH. According to the stoichiometry of the balanced chemical equation, 1 mole of MgCl₂ reacts with 2 moles of KOH to produce 1 mole of Mg(OH)₂ and 2 moles of KCl. Therefore, for 6 moles of MgCl₂ to fully react, you would need 12 moles of KOH.
As only 6 moles of KOH are available, KOH is the limiting reagent because there isn't enough of it to react with all of the 6 moles of MgCl₂; the available amount of KOH dictates the amount of product that can be formed. The excess reactant would be MgCl₂ since not all of it will react. Upon identifying the limiting reagent, it is important to remember that the reaction can only proceed until the limiting reagent is completely consumed, which means in this case, KOH will limit the formation of Magnesium hydroxide (Mg(OH)₂) and potassium chloride (KCl).