Question

In a cycle of copper experiment, a student first reacts a piece of copper metal with nitric acid to produce copper(II) nitrate (Cu(NO3)2) solution. The student then performs various reactions which transform the copper ions into a series of different copper compounds and complexes. Finally, the last reaction reduces the copper ions back to elemental copper metal. Copper atoms are conserved throughout the process. If the initial step of the experiment produces 4.12 mL of 2.41 M Cu(NO3)2, what is the theoretical yield of solid copper (Cu) that can be recovered at the end of the experiment

Answer 1

0.631 grams is the theoretical yield of solid copper (Cu) that can be recovered at the end of the experiment

Step-by-step explanation:

The concentration of the solution is given by :

`[C]=\frac{\text{Moles of compound}}{\text{Volume of solution in Liters}}`

We have:

Concentration of copper (II) nitrate solution =
`[Cu(NO_3)_2]=2.41 M`

The volume of solution = 4.12 mL

1 mL= 0.001 L

`4.12 mL= 4.12* 0.001 L= 0.00412 L`

Moles of copper (II) nitrate in solution = n

`2.41=(n)/(0.00412 L)=0.0099292 mol`

Moles of copper (II) nitrate in solution = 0.0099292 mol

1 Mole of copper(II) nitrate has 1 mole of copper then 0.0099292 moles of copper(II) nitrate will have :

`1* 0.0099292 mol= 0.0099292 \text{ mol of Cu}`

Mass of 0.0099292 moles of copper:

`=0.0099292 mol* 63.55 g/mol=0.63100 g\approx 0.631 g`

This mass of copper present in the solution is the theoretical mass of copper present in the given copper(II) nitrate solution.

0.631 grams is the theoretical yield of solid copper (Cu) that can be recovered at the end of the experiment