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Consider the following reaction. MgCl2(aq)+2NaOH(aq)⟶Mg(OH)2(s)+2NaCl(aq) A 187.0 mL solution of 0.103 M MgCl2 reacts with a 33.58 mL solution of 0.839 M NaOH to produce Mg(OH)2 and NaCl. Identify the limiting reactant.

Respuesta :

Answer:

NaOH is the limiting reactant and MgCl2 is in excess.

Explanation:

let's calculate the mole of each reactant.

number of moles of MgCl2 = [tex]\frac{187}{1000} L* 0.103 mol / L =0.01926 mol\\[/tex]

number of moles of NaOH = [tex]\frac{33.58 }{1000}L * 0.839 mol/L =0.0282 mol[/tex]

as per the reaction , 1 mol of Magnesium choloride reacts with 2 moles of NaOH

hence 0.01926 needs 0.01926x2= 0.03852 moles of NaOH , but there are only 0.0282 moles of NaOH available .

Eduard22sly

The limiting reactant for the reaction between 187.0 mL solution of 0.103 M MgCl₂ and 33.58 mL solution of 0.839 M NaOH is NaOH

We'll begin by calculating the number of mole of MgCl₂ and NaOH present in the solution.

For MgCl₂:

Volume = 187 mL = 187 / 1000 = 0.187 L

Molarity = 0.103 M

Mole of MgCl₂ =?

Mole = Molarity x Volume

Mole of MgCl₂ = 0.103 × 0.187

Mole of MgCl₂ = 0.019 mole

For NaOH:

Volume = 33.58 mL = 33.58 / 1000 = 0.03358 L

Molarity = 0.839 M

Mole of NaOH =?

Mole = Molarity x Volume

Mole of NaOH = 0.839 × 0.03358

Mole of NaOH = 0.028 mole

Finally, we shall determine the limiting reactant. This can be obtained as follow:

MgCl₂ + 2NaOH —> Mg(OH)₂ + 2NaCl

From the balanced equation above,

2 moles of NaOH required 1 mole of MgCl₂.

Therefore, 0.028 mole of NaOH will require = [tex]\frac{0.0282}{2}\\[/tex] = 0.014 mole of MgCl₂

From the calculations made above, we can see that only 0.014 mole of MgCl₂ out of 0.019 mole reacted completely with 0.028 mole of NaOH.

Therefore, NaOH is the limiting reactant and MgCl₂ is the excess reactant.

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