Answer to Question #196018 in Chemistry for Jack

Question #196018

Calculate the mass of the solid produced formed when 1.5 L of 0.50 M NaOH is combined with 2.0 L 0.10 M MgCl2.


1
Expert's answer
2021-05-28T00:56:39-0400

Solution:

Balanced chemical equation:

MgCl2(aq) + 2NaOH(aq) → Mg(OH)2(s) + 2NaCl(aq)

According to the equation above: n(MgCl2) = n(NaOH)/2 = n(Mg(OH)2)


1.5 L of a 0.50 M solution of NaOH contains:

n(NaOH) = concentration × volume = (0.50 mol/L) × (1.5 L) = 0.75 mol NaOH

2.0 L of a 0.10 M solution of MgCl2 contains:

n(MgCl2) = concentration × volume = (0.10 mol/L) × (2.0 L) = 0.20 mol MgCl2


Determine which reagent is limiting:

if NaOH is limiting it would react with (0.75 mol / 2) = 0.375 mol of MgCl2

The initial amount of MgCl2 (0.20 mol) is less than this.

Therefore, MgCl2 is limiting reagent.


Based on the amount of MgCl2 present, the maximum amount of Mg(OH)2 formed is:

n(Mg(OH)2) = n(MgCl2) = 0.20 mol


The molar mass of Mg(OH)2 is 58.32 g/mol.

Therefore,

(0.20 mol Mg(OH)2) × (58.32 g Mg(OH)2 / 1 mol Mg(OH)2) = 11.664 g Mg(OH)2 = 11.7 g Mg(OH)2


Answer: 11.7 grams of the solid produced.

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