Question

if 300ml of 0.4m cuso₄ is reacted with excess naoh, how many grams of cu(oh)₂ precipitate are produced?
cuso₄ + 2naoh → cu(oh)₂ + na₂so₄

Explanation:

Step1: Calculate the moles of \(CuSO_4\)

The formula for molarity \(M=\frac{n}{V}\) (where \(n\) is moles and \(V\) is volume in liters). Given \(V = 300\space mL=0.3\space L\) and \(M = 0.4\space M\).
\(n(CuSO_4)=M\times V=0.4\times0.3 = 0.12\space mol\)

Step2: Determine the moles of \(Cu(OH)_2\)

From the balanced chemical equation \(CuSO_4+2NaOH
ightarrow Cu(OH)_2 + Na_2SO_4\), the mole ratio of \(CuSO_4\) to \(Cu(OH)_2\) is \(1:1\). So \(n(Cu(OH)_2)=n(CuSO_4) = 0.12\space mol\)

Step3: Calculate the mass of \(Cu(OH)_2\)

The molar mass of \(Cu(OH)_2\) is \(M_{Cu(OH)_2}=64+(16 + 1)\times2=64 + 34=98\space g/mol\).
Using the formula \(m=n\times M\), \(m(Cu(OH)_2)=0.12\times98 = 11.76\space g\)

Answer:

\(11.76\space g\)