This experiment involved mixing aqueous solutions of Cu(NO3)2 (Copper nitrate) a

Question

This experiment involved mixing aqueous solutions of Cu(NO3)2 (Copper nitrate) and KI (potassium iodide)

2 Cu(NO3)2 (aq) + 4 KI (aq) ---- > 2 CuI (s) + I2 (aq) + 4KNO3 (aq)

You have two test tubes labeled A and B. Both test tubes have the same solution in them. The solution is 2.0 mL of 0.50 M Cu(NO3)2 and 5.0 mL of 0.50 M KI. After centrifuging the test tubes for 5 minutes the preciptate (CuI) settles to the bottom. The supernatant is removed from each test tube and then the precipitate is dried. Once it is dry, the precipitate is weighed in each test tube. Test tube A ended up with a precipitate weighing 0.3 grams and test tube B had a precipitate weighing 0.33 grams.

Given this information, calculate the following:

1.) average moles of product (CuI)

2.) average moles of Cu(NO3)2

3.) average moles of KI

4.) theoretical moles of product based Cu(NO3)2

5.) theoretical moles based on KI

6.) percent error in moles of CuI

Explanation / Answer

1.

In test tube A, mole of CuI = 0.3 /190.48 = 1.57 x 10^-3 mol

In test tube B, mole of CuI = 0.33 /190.48 = 1.37 x 10^-3 mol

So, average of moles = ( 1.57 x 10^-3+ 1.37 x 10^-3 ) / 2 = 4.47 x 10^-3 mol

2.

In test tube A, mole of Cu(NO3)2= 0.50 x 0.002 = 0.001 mol

In test tube B, mole of Cu(NO3)2= 0.50 x 0.002 = 0.001 mol

So, average of moles = ( 0.001+ 0.001 ) / 2 = 0.001mol

3.

In test tube A, mole of KI = 0.50 x 0.005 = 0.0025 mol

In test tube B, mole of KI= 0.50 x 0.005 = 0.0025 mol

So, average of moles = ( 0.0025+ 0.0025 ) / 2 = 0.0025 mol

4.

theoretical moles of product based Cu(NO3)2 is 1 moles

5.

theoretical moles based on KI is 0.5 mol

6.

percent error in moles of CuI = (Expeected value - true value / true value) x 100 = (1 -0.00447/1) x 100 = 99.6 %

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