In the reaction MgCl2 + 2KOH Mg (OH) 2 + 2KCl, if 6 moles MgCl2 are added to 6 moles KOH, which is the limiting reagent?

Question

Uriel

Chemistry

7 June, 17:04

In the reaction MgCl2 + 2KOH Mg (OH) 2 + 2KCl, if 6 moles MgCl2 are added to 6 moles KOH, which is the limiting reagent?

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Answers (1)

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Braiden Phillips · Answer 1

One way to find the limiting reagent is to convert both reactants into one of the products.

6 moles MgCl2 (1 mol Mg (OH) 2 / 1 mole MgCl2) = 6 mol Mg (OH) 2

6 moles KOH (1 mol Mg (OH) 2 / 2 mol KOH) = 3 mol Mg (OH) 2

since KOH produce the least amount of moles, KOH is the limiting reagent