Topics: Mole, Kilogram, Stoichiometry Pages: 2 (495 words) Published: February 14, 2013
Chemistry Help! Could someone please help me with these chemistry problems. I am so lost. Thanks!? You mix 4.0 mL of 0.50 M Cu(NO3)2 with 3.0 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 0.11 g, what is the percent recovery of the precipitate? 3 years ago Report Abuse

Best Answer - Chosen by Voters

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

Total Moles of Cu(NO3)2 in solution = 0.50 mol/L * (4.0mL*1L/1000mL) = 0.002 mol

Total Moles of KI in solution = 0.50 mol/L * (3.0mL*1L/1000mL) = 0.0015 mol

Mole ratio of Cu(NO3)2 to CuI = 2:2 = 1:1

Mole ratio of KI to CuI = 4:2 = 2:1

Therefore, more KI is needed making it your limiting reactant.

This means all 0.0015 moles of KI will react with Cu(NO3)2. By comparing mole ration of KI to Cu(NO3)2, you find that Cu(NO3)2:KI is 2:4 or 1:2. Hence:

0.0015 mol of KI * (1 mole of Cu(NO3)2 / 2 moles of KI) = 0.0075 moles of Cu(NO3)2

For every 2 moles of KI you get 1 mole of CuI, so:

0.0015 mol KI * (1 mole of CuI / 2 moles of KI) = 0.00075 moles of CuI formed

You can also check this by comparing Cu(NO3)2 to CuI via mole ratio and you should get the same amount.

Now that you know the moles of CuI, use the molar mass of CuI (190.45g/mol) to get mass of CuI:

0.00075 moles of CuI * (190.45g/mol) = 0.1428375 grams = 0.14 grams b/c of sig figs

Since only 0.00075 moles of Cu(NO3)2 reacted with all of 0.0015 moles of KI, it is evident that the moles of KI used was 0.0015 moles and the moles of Cu(NO3)2 used was 0.00075 moles.

The moles expected from Cu(NO3)2 in solution would be 0.002 moles because it is a 1:1 ration between Cu(NO3)2 and CuI.

The moles expected from KI in solution would be...
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