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

Respuesta :

JavierCollante
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
Dejanras

Answer is: potassium hydroxide (KOH).

Balanced chemical reaction: MgCl₂ + 2KOH → Mg(OH)₂ + 2KCl.

n(MgCl₂) = 6 mol; amount of magnesium chloride.

n(KOH) = 6 mol; amount of potassium hydroxide.

From balanced chemical reaction: n(MgCl₂) : n(KOH) = 1 : 2.

6 mol : n(KOH) = 1 : 2.

n(KOH) = 6 mol · 2.

n(KOH) = 12 mol; potassium hydroxide is limiting reagent, because we need 12 moles of it for chemical reaction and there is only 6 moles.

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