Limiting and Excess Reagents Calculating Molarity: What is the formula weight of

Question

Limiting and Excess Reagents Calculating Molarity: What is the formula weight of Cul? What quantity of Cul is present in 226 mL of a 0.20 M solution? What is the formula weight of Cu(NO_3)_2? What mass of Cu(NO_3)_2 is present in 151 mL of a 0.60 M solution? What is the formula weight of KI? What mass of KI is needed to add to a 1000 mL volumetric flask to produce a solution that is 0.40 M? What is the formula weight of I_2? What is the molarity of a solution produced by dissolving 41 g of I_2 in a 500 mL volumetric flask? Limiting Reagent Calculations: You mix 6.0 mL of 0.50 M Cu(NO_3)_2 with 1.0 mL of 0.50 M KI, and collect and dry the Cul precipitate. Calculate the following: Moles of Cu(NO_3)_2 used: Moles of KI used. Moles of Cul precipitate expected from Cu(NO_3)_2: Moles of Cul precipitate expected from KI: What mass of Cul do you expect to obtain from this reaction? If the actual mass of the precipitate you recover is 0.038 g, what is the percent recovery of the precipitate?

Explanation / Answer

2. The formula weight of CuI IS 190.45 gm/mol.

As, 1000 ml 1(M) CuI = 190.45 gm

So, 226 ml 0.2 (M) CuI = 168.53 gm

The formula weight of Cu (NO3)2 = 187.56 gm/mol

As 1000ml 1M Cu(NO3)2 = 187.56 gm

151 ml 0.6 (M) Cu(NO3)2 = 745.27 gm

Formula weight of KI = 166 gm

1000 ml 1(M) KI = 166 gm

1000ml 0.4 (M) = 66.4 gm

Formula Weight of I2 = 253.81 gm/mol

In 1000 ml if 253.81 gm iodine is present, molarity is 1 M

Then the molarity of aforesaid solution of iodine is 12.38 M

2. 0.003 mole of Cu(NO3)2 is used

0.0005 mole of KI is used

2KI + Cu( NO3)2 = CuI2 + 2KNO3

So from 0.003 mol of copper nitrate 0.003 mole of CuI2 will expected to produce.

and from 0.0005 mole of KI 0.000025 mole of CuI2 will expected to produce.

Note: Please check the question, THERE IS ERROR THAT IS NOT CuI WILL PPT OUT , IT WILL BE CuI2

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