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.
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 %
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