In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture: 5.00 mL 3.40 M acetone + 10.0 mL 1.50 M HCI + 10.0 mL 0.00400 M I, + 25.0 mL H20 a. How many moles of acetone were in the reaction mixture? Recall that, for a component A, moles of A = [A] x V where [A] is the molarity of A and V is the volume in liters of the solution of A that was used. 0.017 moles b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50.0 mL, 0.0500 L, and the number of moles of acetone was found in Part (a). moles of A [A] %3D V of soln in liters 0.34 M acetone c. How could you double the molarity of acetone in the reaction mixture, keeping the total volume at 50 mL and keeping the same concentrations of Ht ion and I2 as in the original mixture? Use 6.80 mL acetone and 50 mL water, keeping the volumes of HCl and I2 the same.

Chemistry: The Molecular Science
5th Edition
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Author:John W. Moore, Conrad L. Stanitski
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Chapter12: Chemical Equilibrium
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In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture:
5.00 mL 3.40 M acetone + 10.0 mL 1.50 M HCI + 10.0 mL 0.00400 M I, + 25.0 mL H20
a. How many moles of acetone were in the reaction mixture? Recall that, for a component A,
moles of A
[A] x V
where A is the molarity of A and V is the volume in liters of the solution of A that was used.
0.017
moles
b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50.0 mL, 0.0500 L, and the number of moles of
acetone was found in Part (a).
moles of A
[A]
%3D
V of soln in liters
0.34
M acetone
c. How could you double the molarity of acetone in the reaction mixture, keeping the total volume at 50 mL and keeping the same concentrations
of HT ion and 12 as in the original mixture?
Use 6.80
mL acetone and 50
mL water, keeping the volumes of HCl and I2 the same.
Transcribed Image Text:In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture: 5.00 mL 3.40 M acetone + 10.0 mL 1.50 M HCI + 10.0 mL 0.00400 M I, + 25.0 mL H20 a. How many moles of acetone were in the reaction mixture? Recall that, for a component A, moles of A [A] x V where A is the molarity of A and V is the volume in liters of the solution of A that was used. 0.017 moles b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50.0 mL, 0.0500 L, and the number of moles of acetone was found in Part (a). moles of A [A] %3D V of soln in liters 0.34 M acetone c. How could you double the molarity of acetone in the reaction mixture, keeping the total volume at 50 mL and keeping the same concentrations of HT ion and 12 as in the original mixture? Use 6.80 mL acetone and 50 mL water, keeping the volumes of HCl and I2 the same.
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