Question

Under a certain set of conditions, the reaction of copper(II) nitrate with sodium hydroxide produces copper(II) hydroxide in 96.5% yield. What mass of copper(II) nitrate would be required to produce 11.4 mg Cu(OH)2 under these conditions?

Answer 1

The balanced reaction is

Cu(NO3)2 (aq) + 2NaOH (aq) = Cu(OH)2 (s) + 2NaNO3 (aq)

% yield = actual yield x 100 / theoretical yield

Theoretical yield = actual yield /%yield

= 11.4/0.965 = 11.813 mg x 1g/1000 mg = 0.011813 g

Moles of Cu(OH)2 = theoretical yield /molecular weight

= 0.011813 g / 97.561 g/mol

= 0.00012108 moles

Now determine the moles of Cu(NO3)2 required to produce 0.00012108 moles of Cu(OH)2

From the stoichiometry of the reaction

Moles of Cu(NO3)2 = Moles of Cu(OH)2 = 0.00012108

Mass of Cu(NO3)2 = moles x molecular weight

= 0.00012108 moles x 187.56 g/mol

= 0.0227 g x 1000mg/g

= 22.7 mg