Question

A copper cycle experiment takes copper atoms through reactions that produce copper compounds and complexes one after the other, finally producing elemental copper. Copper atoms are conserved throughout the process. Given that a student begins with 9.29 mL of a 0.14 M Cu(NO3)2 solution, how much copper should be isolated at the end of the cycle?

Answer 1

To find the amount of copper to be isolated, calculate the initial moles of Cu(NO3)2 from the given volume and concentration, and then convert that amount to grams using copper's atomic weight. Approximately 82.7 mg of copper should be recovered at the end of the copper cycle experiment.

Step-by-step explanation:

To determine how much copper should be isolated at the end of the experiment, we start by calculating the initial amount of copper in moles present in a 9.29 mL of a 0.14 M Cu(NO3)2 solution. The molarity (M) indicates moles of solute per liter of solution, so:

Initial moles of Cu(NO3)2 = Volume (L) × Molarity (M)

= 0.00929 L × 0.14 mol/L

= 0.0013006 mol

Since copper (II) nitrate, Cu(NO3)2, contains one copper atom per formula unit, the moles of copper will be the same as the moles of the copper nitrate: 0.0013006 mol. The atomic weight of copper is approximately 63.55 g/mol, so:

Mass of copper = Moles of copper × Atomic weight of copper

= 0.0013006 mol × 63.55 g/mol

= 0.0827 g or 82.7 mg

Therefore, assuming 100% efficiency and no loss of copper through the cycle, the student should isolate approximately 82.7 mg of elemental copper by the end of the copper cycle experiment.

Answer 2

Answer: 0.082 g

Step-by-step explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

`Molarity=(n* 1000)/(V_s)`

where,

Molarity = 0.14 M

n= moles of solute = ?

`V_s` = volume of solution in ml = 9.29 ml

`0.14=(n* 1000)/(9.29)`

`n=(0.14* 9.29)/(1000)=1.3* 10^(-3)moles`

Now as 1 mole of
`Cu(NO_3)_2` contains 1 mole of copper

`1.3* 10^(-3)moles` moles of
`Cu(NO_3)_2` contains =
`(1)/(1)* 1.3* 10^(-3)=1.3* 10^(-3)moles` of copper

Mass of copper =
`moles* {\text {molar mass}}=1.3* 10^(-3)* 63.5=0.082g`

Thus 0.082 g of copper should be isolated at the end of the cycle.