What is the minimum volume of CuNO3 should be used in this reaction?

Question

The following reaction is known to have a 65.0% yield of iron.

3 Cu(NO₃)_2(aq) + 2 Fe -> 2 Fe(NO₃)_2(aq) + 3 Cu

To produce 25.0 g of copper, what is the minimum volume of CuNO3 should be used in this reaction, assuming that iron is in excess?

Answer 1

3 Cu(NO₃)_2(aq) + 2 Fe -> 2 Fe(NO₃)_2(aq) + 3 Cu ... balanced equation

65.0% of what mass of Cu = 25.0 g Cu?

0.650 x = 25.0 g and x = 38.5 g Cu

This is the theoretical yield of Cu such that at 65% yield, you will have 25.0 g Cu.

From this, and the stoichiometry of the balanced equation, you can calculate the moles of Cu(NO3)2 needed. Then, IF YOU KNEW THE CONCENTRATION OF THE Cu(NO3)2, you could find the volume.

38.5 g Cu x 1 mol Cu/63.5 g x 3 mol Cu (NO3)2 / 3 mol Cu = 0.606 moles Cu(NO3)2 needed

From here you'd have to know the concentration of the Cu(NO3)2 to find the volume needed