Answer to Question #324500 in General Chemistry for Ivan Ackerman

Question

Consider the following reaction.MgCl

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

A 127.0 mL solution of 0.132 M MgCl2 reacts with a 33.22 mL solution of 0.664 M NaOH  to produce Mg(OH)2 and NaCl.

Caclulate the mass of Mg(OH)2 that can be produced.

The actual mass of Mg(OH)2 isolated was 0.501 g. Calculate the percent yield of Mg(OH)2

.

Answer 1

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

n(MgCl₂) = C(sol) × V(sol) = 0.132 mol/L × 0.127 L = 0.016764 mol

n(NaOH) = 0.664 mol/L × 0.03322 L = 0.022058 mol

According chemical reaction

1 mol of MgCl₂ reacts with 2 moles of NaOH

NaOH is limiting reactant

2 mol NaOH - 1 mol Mg(OH)₂

0.022058 mol NaOH - x mol Mg(OH)₂

x = 0.011029 mol

m(Mg(OH)₂) = n(Mg(OH)₂) × Mr(Mg(OH)₂) = 0.011029 mol × 58.3197 g/mol = 0.6432 g

Percent yield = m_practical / m_theoretical × 100% = 0.501 g / 0.6432 g × 100% = 77.89%