Question

In the reaction MgCl2 + 2KOH Mg(OH)2 + 2KCl, if 3 moles MgCl2 are added to 4 moles KOH, what determines how much Mg(OH)2 is made?

A. The amount of MgCl2 + KOH
B. The amount of KCl
C. The amount of MgCl2
D. The amount of KOH

Answer 1

The amount of Mg(OH)2 formed in the reaction is determined by the limiting reactant, which in this case is KOH because there is not enough KOH to react with all the MgCl2 present.

Step-by-step explanation:

In the reaction MgCl2 + 2KOH → Mg(OH)2 + 2KCl, the amount of Mg(OH)2 that can be made is determined by the limiting reactant. The limiting reactant is the reactant that is completely used up first, limiting the amount of product that can be formed. To determine the limiting reactant, we compare the mole ratio of the reactants used in the reaction to the mole ratio of the reactants we have.

In this case, the balanced chemical equation indicates that 1 mole of MgCl2 reacts with 2 moles of KOH. Since 3 moles of MgCl2 are added to 4 moles of KOH, KOH is the limiting reactant because there is not enough KOH to react with all the MgCl2: 3 moles of MgCl2 would require 6 moles of KOH for the reaction to go to completion. Thus, the amount of Mg(OH)2 formed is determined by the amount of KOH present.

Answer 2

The correct option is C. The amount of MgCl2. we know this because no matter how much you increase KOH, if you dont increase Mgcl2, the amount of Mg(OH)2 remains the same. Hope this works for you