In a reaction involving the iodination of acetone, the following volumes were us

Question

In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture: 10 mL 4.0 M acetone + 10 mL. 1.0 M HCI+ 10 mL 0.0050 M I,+20 mL H,0 How many moles of acetone were in the reaction mixture? Recall that, for a component A, no. moles A M, x V, where M, is the molarity of A and V is the volume in liters of the solution of A that was used. a. moles acetone b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50 mL 0.050 L, and the number of moles of acetone was found in Part (a). Again, no. moles A A Vof soln. in liters M acetone How could you double the molarity of the acetone in the reaction mixture, keeping the total volume c. at 50 mL and keeping the same concentrations of H' ion and I, as in the original mixture? Using the reaction mixture in Problem 1, a student found that it took 230 seconds for the color of the L, to disappear. a. What was the rate of the reaction? Hint: First find the initial concentration of I, in the reaction mixture, (12o Then use Equation 5. rate = b. Given the rate from Part (a), and the initial concentrations of acetone, H ion, and I2 in the reaction mixture, write Equation 3 as it would apply to the mixture. rate = c. What are the unknowns that remain in the equation in Part (b)?

Explanation / Answer

a) no. of moles = M*V

= 4*10ml = 4 mmol or 0.04 moles

b) molarity of acetone = 4mmol/50mL = 0.08 M

c) As number of moles is directly proportional to molarity, doubling the moles will double to molarity, provided total volume is constant.

(note: full snapshot of the question is not provided, the above mentioned equation 5 is missing/ not highlighted.)

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