Question

Consider the following reaction. MgCl2(aq)+2NaOH(aq)⟶Mg(OH)2(s)+2NaCl(aq) A 166.0 mL solution of 0.381 M MgCl2 reacts with a 47.33 mL solution of 0.568 M NaOH to produce Mg(OH)2 and NaCl. Identify the limiting reactant. NaOH Mg(OH)2 MgCl2 NaCl Caclulate the mass of Mg(OH)2 that can be produced. The actual mass of Mg(OH)2 isolated was 0.559 g. Calculate the percent yield of Mg(OH)2.

Answer 1

1)

volume of MgCl2, V = 1.66*10^2 mL

= 0.166 L

use:

number of mol in MgCl2,

n = Molarity * Volume

= 0.381*0.166

= 6.325*10^-2 mol

volume of NaOH, V = 47.33 mL

= 4.733*10^-2 L

use:

number of mol in NaOH,

n = Molarity * Volume

= 0.568*4.733*10^-2

= 2.688*10^-2 mol

Balanced chemical equation is:

MgCl2 + 2 NaOH ---> Mg(OH)2 + 2 NaCl

1 mol of MgCl2 reacts with 2 mol of NaOH

for 6.325*10^-2 mol of MgCl2, 0.1265 mol of NaOH is required

But we have 2.688*10^-2 mol of NaOH

so, NaOH is limiting reagent

Answer: NaOH

2)

we will use NaOH in further calculation

Molar mass of Mg(OH)2,

MM = 1*MM(Mg) + 2*MM(O) + 2*MM(H)

= 1*24.31 + 2*16.0 + 2*1.008

= 58.326 g/mol

According to balanced equation

mol of Mg(OH)2 formed = (1/2)* moles of NaOH

= (1/2)*2.688*10^-2

= 1.344*10^-2 mol

use:

mass of Mg(OH)2 = number of mol * molar mass

= 1.344*10^-2*58.33

= 0.784 g

Answer: 0.784 g

3)

% yield = actual mass*100/theoretical mass

= 0.559*100/0.784

= 71.3%

Answer: 71.3 %