Limiting Reagent Calculations?

Question

You mix 5.0 mL of 0.50 M Cu(NO3)2 with 2.0 mL of 0.50 M KI, and collect and dry the CuI precipitate. Calculate the following:


Moles of Cu(NO3)2 used:


Moles of KI used:


Moles of CuI precipitate expected from Cu(NO3)2.


Moles of CuI precipitate expected from KI.

What mass of CuI do you expect to obtain from this reaction?

If the actual mass of the precipitate you recover is 0.081 g, what is the percent recovery of the precipitate?

Answer 1

The reaction takes two steps. The first step is:
Cu(NO3)2 + 2KI → CuI2 + 2KNO3
The CuI2 immediately decomposes to iodine and copper(I) iodide, releasing I2:
2CuI2 → 2CuI + I2
Combine these two steps together, we have:
2Cu(NO3)2 + 4KI → 2CuI + I2 + 4KNO3
1) Moles of Cu(NO3)2 used:
5.0 mL of 0.50 M Cu(NO3)2 is 0.0050*0.50 = 0.0025 Mol of Cu(NO3)2
2) Moles of KI used:
2.0 mL of 0.50 M KI is 0.0020*0.50 = 0.0010 Mol of KI.
3) Moles of CuI precipitate expected from Cu(NO3)2: 0.0025 Mol
4) Moles of CuI precipitate expected from KI: 0.00050 Mol < only half of the molar number of KI since half of the iodine would be oxidized to I2>
5) What mass of CuI do you expect to obtain from this reaction: CuI is poorly soluble in water (0.00042 g/L at 25 °C). We will omit this solubility since the reaction temperature is not specified. Now KI is the limiting agent, thus only a maximum of 0.00050 Mol of CuI can be formed, and that is 190.45*0.00050 = 0.095(gram)
6) The percent recovery of the precipitate is: 0.081/0.095 = 85%.

Answer 2

Don't Know