In a reaction involving the iodination of acetone. the following volume were use

Question

In a reaction involving the iodination of acetone. the following volume were used to make up the reaction mixture: 10 0 ml 4.0 M acetone + 10.0 ml. 1.0 M HCL + 10.0 ml 0.0050 M l2 + 20.0 ml H2O How many moles of acetone Here in the reaction mixture? Recall that, for a component A. no mole A = MA x V1 where MA is the morality of A and 1' the volume in liters of the solution of A that was used. When was the morality of acetone in the reaction mixture? The volume of the mixture was 50 ml, 0.050 liter, and the number of moles of acetone was found in Part a. How could you double the morality of the acetone in the reaction mixture, keeping the total volume at 50 ml and keeping the same concentrations of H+ ion and I2 as in the original mixture?

Explanation / Answer

1.   (a). Moles of Acetone = (10/1000)(4) = 0.04 moles

(b). Molarity of acetone = no. of moles of acetone / volume of solution = 0.04/0.05 = 0.8 M

(c). We can double the concentration of acetone by increasing the volume of acetone from 10ml to 20 ml and by decreasing the volume of water fropm 20ml to 10 ml.

2. (a) Initial moles of I2 = (10/1000)(0.0050) = 0.00005 moles

Initial concentration = 0.00005/0.05 = 0.001 M

Final conc. = 0 M

Change in conc. = 0.001 M

Rate of reaction= 0.001/310 = 3.22 X 10-6 moles/L.sec

(b). Initial conc of HCl= (0.01)(1.0)/0.05 = 0.2 M

Initial conc of I2 = 0.001 M

Initial conc of acetone = 0.8 M

R = k[Acetone][HCl][I2]

(c) Unknowns remained are : k , [Acetone] , [HCl] and [I2]

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