Question

determine the mass of mg(oh)2 formed when 250ml of 0.3m mgcl2 reacts completely with an excess naoh. mgcl2 + 2naoh → mg(oh)2 + 2nacl

Explanation:

Step1: Calculate the number of moles of \(MgCl_2\)

The formula for molarity \(M=\frac{n}{V}\) (where \(n\) is the number of moles and \(V\) is the volume in liters). Given \(M = 0.3\space M\) and \(V=250\space mL=0.25\space L\).

$$n(MgCl_2)=M\times V=0.3\times0.25 = 0.075\space mol$$

Step2: Use the stoichiometry of the reaction

From the balanced equation \(MgCl_2 + 2NaOH
ightarrow Mg(OH)_2+2NaCl\), the mole ratio of \(MgCl_2\) to \(Mg(OH)_2\) is \(1:1\). So, \(n(Mg(OH)_2)=n(MgCl_2) = 0.075\space mol\)

Step3: Calculate the molar mass of \(Mg(OH)_2\)

The molar mass of \(Mg(OH)_2\): \(M = 24+(16 + 1)\times2=24 + 34=58\space g/mol\)

Step4: Calculate the mass of \(Mg(OH)_2\)

Using the formula \(m=n\times M\), where \(n = 0.075\space mol\) and \(M = 58\space g/mol\)

$$m(Mg(OH)_2)=0.075\times58=4.35\space g$$

Answer:

\(4.35\space g\)